right triangles and trigonometry homework 4

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Unit 1: Right triangles & trigonometry

About this unit, ratios in right triangles.

• Getting ready for right triangles and trigonometry (Opens a modal)
• Hypotenuse, opposite, and adjacent (Opens a modal)
• Side ratios in right triangles as a function of the angles (Opens a modal)
• Using similarity to estimate ratio between side lengths (Opens a modal)
• Using right triangle ratios to approximate angle measure (Opens a modal)
• Right triangles & trigonometry: FAQ (Opens a modal)
• Use ratios in right triangles Get 3 of 4 questions to level up!

Introduction to the trigonometric ratios

• Triangle similarity & the trigonometric ratios (Opens a modal)
• Trigonometric ratios in right triangles (Opens a modal)
• Trigonometric ratios in right triangles Get 3 of 4 questions to level up!

Solving for a side in a right triangle using the trigonometric ratios

• Solving for a side in right triangles with trigonometry (Opens a modal)
• Solve for a side in right triangles Get 3 of 4 questions to level up!

Solving for an angle in a right triangle using the trigonometric ratios

• Intro to inverse trig functions (Opens a modal)
• Solve for an angle in right triangles Get 3 of 4 questions to level up!

Sine and cosine of complementary angles

• Intro to the Pythagorean trig identity (Opens a modal)
• Sine & cosine of complementary angles (Opens a modal)
• Using complementary angles (Opens a modal)
• Trig word problem: complementary angles (Opens a modal)
• Trig challenge problem: trig values & side ratios (Opens a modal)
• Trig ratios of special triangles (Opens a modal)
• Relate ratios in right triangles Get 3 of 4 questions to level up!

Modeling with right triangles

• Right triangle word problem (Opens a modal)
• Angles of elevation and depression (Opens a modal)
• Right triangle trigonometry review (Opens a modal)
• Right triangle trigonometry word problems Get 3 of 4 questions to level up!

The reciprocal trigonometric ratios

• Reciprocal trig ratios (Opens a modal)
• Finding reciprocal trig ratios (Opens a modal)
• Using reciprocal trig ratios (Opens a modal)
• Trigonometric ratios review (Opens a modal)
• Reciprocal trig ratios Get 5 of 7 questions to level up!

Chapter 4: Trigonometric Functions

Section 4.3: right triangle trigonometry, learning outcomes.

• Use right triangles to evaluate trigonometric functions.
• Find function values for 30° (π/6),   45° (π/4), and 60° (π/3).
• Use cofunctions of complementary angles.
• Use the definitions of trigonometric functions of any angle.
• Use right triangle trigonometry to solve applied problems.

Using Right Triangles to Evaluate Trigonometric Functions

In earlier sections, we used a unit circle to define the trigonometric functions . In this section, we will extend those definitions so that we can apply them to right triangles. The value of the sine or cosine function of [latex]t[/latex] is its value at [latex]t[/latex] radians. First, we need to create our right triangle. Figure 1 shows a point on a unit circle of radius 1. If we drop a vertical line segment from the point [latex]\left(x,y\right)\\[/latex] to the x -axis, we have a right triangle whose vertical side has length [latex]y[/latex] and whose horizontal side has length [latex]x[/latex]. We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle.

Likewise, we know

These ratios still apply to the sides of a right triangle when no unit circle is involved and when the triangle is not in standard position and is not being graphed using [latex]\left(x,y\right)[/latex] coordinates. To be able to use these ratios freely, we will give the sides more general names: Instead of [latex]x[/latex], we will call the side between the given angle and the right angle the adjacent side to angle [latex]t[/latex]. (Adjacent means “next to.”) Instead of [latex]y[/latex], we will call the side most distant from the given angle the opposite side from angle [latex]t[/latex]. And instead of [latex]1[/latex], we will call the side of a right triangle opposite the right angle the hypotenuse . These sides are labeled in Figure 2.

Figure 2.  The sides of a right triangle in relation to angle [latex]t[/latex].

Understanding Right Triangle Relationships

Given a right triangle with an acute angle of [latex]t[/latex],

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “ S ine is o pposite over h ypotenuse, C osine is a djacent over h ypotenuse, T angent is o pposite over a djacent.”

How To: Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle.

• Find the sine as the ratio of the opposite side to the hypotenuse
• Find the cosine as the ratio of the adjacent side to the hypotenuse.
• Find the tangent is the ratio of the opposite side to the adjacent side.

Example 1: Evaluating a Trigonometric Function of a Right Triangle

Given the triangle shown in Figure 3, find the value of [latex]\cos \alpha[/latex].

The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so:

[latex]\begin{align}\cos \left(\alpha \right)=\frac{\text{adjacent}}{\text{hypotenuse}} =\frac{15}{17} \end{align}[/latex]

Given the triangle shown in Figure 4, find the value of [latex]\text{sin}t[/latex].

[latex]\frac{7}{25}[/latex]

Relating Angles and Their Functions

When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure 5. The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

Figure 5. The side adjacent to one angle is opposite the other.

We will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.

How To: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.

• If needed, draw the right triangle and label the angle provided.
• Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
• sine as the ratio of the opposite side to the hypotenuse
• cosine as the ratio of the adjacent side to the hypotenuse
• tangent as the ratio of the opposite side to the adjacent side
• secant as the ratio of the hypotenuse to the adjacent side
• cosecant as the ratio of the hypotenuse to the opposite side
• cotangent as the ratio of the adjacent side to the opposite side

Example 2: Evaluating Trigonometric Functions of Angles Not in Standard Position

Using the triangle shown in Figure 6, evaluate [latex]\sin \alpha[/latex], [latex]\cos \alpha[/latex], [latex]\tan \alpha[/latex], [latex]\sec \alpha[/latex], [latex]\csc \alpha [/latex], and [latex]\cot \alpha[/latex].

[latex]\begin{align}&\sin \alpha =\frac{\text{opposite }\alpha }{\text{hypotenuse}}=\frac{4}{5}\\ &\cos \alpha =\frac{\text{adjacent to }\alpha }{\text{hypotenuse}}=\frac{3}{5} \\ &\tan \alpha =\frac{\text{opposite }\alpha }{\text{adjacent to }\alpha }=\frac{4}{3} \\ &\sec \alpha =\frac{\text{hypotenuse}}{\text{adjacent to }\alpha }=\frac{5}{3} \\ &\csc \alpha =\frac{\text{hypotenuse}}{\text{opposite }\alpha }=\frac{5}{4} \\ &\cot \alpha =\frac{\text{adjacent to }\alpha }{\text{opposite }\alpha }=\frac{3}{4} \end{align}[/latex]

Using the triangle shown in Figure 7, evaluate [latex]\sin t[/latex], [latex]\cos t[/latex], [latex]\tan t[/latex], [latex]\sec t[/latex], [latex]\csc t[/latex], and [latex]\cot t[/latex].

[latex]\begin{align}&\sin t=\frac{33}{65},\cos t=\frac{56}{65},\tan t=\frac{33}{56}, \\ &\sec t=\frac{65}{56},\csc t=\frac{65}{33},\cot t=\frac{56}{33} \end{align}[/latex]

Finding Trigonometric Functions of Special Angles Using Side Lengths

We have already discussed the trigonometric functions as they relate to the special angles on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of [latex]30^\circ [/latex], [latex]60^\circ [/latex], and [latex]45^\circ[/latex], however, remember that when dealing with right triangles, we are limited to angles between [latex]0^\circ \text{ and } 90^\circ[/latex].

Suppose we have a [latex]30^\circ ,60^\circ ,90^\circ [/latex] triangle, which can also be described as a [latex]\frac{\pi }{6}, \frac{\pi }{3},\frac{\pi }{2}[/latex] triangle. The sides have lengths in the relation [latex]s,\sqrt{3}s,2s[/latex]. The sides of a [latex]45^\circ ,45^\circ ,90^\circ [/latex] triangle, which can also be described as a [latex]\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{2}[/latex] triangle, have lengths in the relation [latex]s,s,\sqrt{2}s[/latex]. These relations are shown in Figure 8.

Figure 8. Side lengths of special triangles

We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.

How To: Given trigonometric functions of a special angle, evaluate using side lengths.

• Use the side lengths shown in Figure 8 for the special angle you wish to evaluate.
• Use the ratio of side lengths appropriate to the function you wish to evaluate.

Example 3: Evaluating Trigonometric Functions of Special Angles Using Side Lengths

Find the exact value of the trigonometric functions of [latex]\frac{\pi }{3}[/latex], using side lengths.

[latex]\begin{align}&\sin \left(\frac{\pi }{3}\right)=\frac{\text{opp}}{\text{hyp}}=\frac{\sqrt{3}s}{2s}=\frac{\sqrt{3}}{2}\\ &\cos \left(\frac{\pi }{3}\right)=\frac{\text{adj}}{\text{hyp}}=\frac{s}{2s}=\frac{1}{2}\\ &\tan \left(\frac{\pi }{3}\right)=\frac{\text{opp}}{\text{adj}}=\frac{\sqrt{3}s}{s}=\sqrt{3}\\ &\sec \left(\frac{\pi }{3}\right)=\frac{\text{hyp}}{\text{adj}}=\frac{2s}{s}=2\\ &\csc \left(\frac{\pi }{3}\right)=\frac{\text{hyp}}{\text{opp}}=\frac{2s}{\sqrt{3}s}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3} \\ &\cot \left(\frac{\pi }{3}\right)=\frac{\text{adj}}{\text{opp}}=\frac{s}{\sqrt{3}s}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3} \end{align}[/latex]

Find the exact value of the trigonometric functions of [latex]\frac{\pi }{4}[/latex], using side lengths.

[latex]\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2},\cos \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2},\tan \left(\frac{\pi }{4}\right)=1[/latex], [latex]\sec \left(\frac{\pi }{4}\right)=\sqrt{2},\csc\left(\frac{\pi }{4}\right)=\sqrt{2},\cot \left(\frac{\pi }{4}\right)=1[/latex]

By looking at all six trig functions and their special angles, we can organize it all in one place.

Now that we have this table, we can use it to find the exact values of trigonometric expressions.

Example 4: Evaluating Trigonometric Functions of Special Angles

Find the exact value of the trigonometric function [latex]\frac{\sin\left(\frac{\pi}{6}\right)}{1+\cos^{2}\left(\frac{\pi}{6}\right)}[/latex] using the table above.

[latex]\begin{gathered} & \frac{\frac{1}{2}}{1+\left(\frac{\sqrt{3}}{2}\right)^{2}} && \text{Put in the values from the table. Note the squaring of cosine.} \\ &\frac{\frac{1}{2}}{1+\frac{3}{4}} && \text{Square the fraction.} \\ &\frac{\frac{1}{2}}{\frac{7}{4}} && \text{Get common denominators.} \\ &\left(\frac{1}{2}\right)\left({\frac{4}{7}}\right) && \text{Get common denominators.} \\ &\frac{2}{7} && \text{Simplify}\end{gathered}[/latex]

Find the exact value of the trigonometric function [latex]\sec^{2}\left(45^\circ\right)+\tan^{2}\left(60^\circ\right)[/latex] using the table above.

Using Equal Cofunction of Complements

If we look at the table above, we will notice a pattern. In a right triangle with angles of [latex]\frac{\pi }{6}[/latex] and [latex]\frac{\pi }{3}[/latex], we see that the sine of [latex]\frac{\pi }{3}[/latex], namely [latex]\frac{\sqrt{3}}{2}[/latex], is also the cosine of [latex]\frac{\pi }{6}[/latex], while the sine of [latex]\frac{\pi }{6}[/latex], namely [latex]\frac{1}{2}[/latex], is also the cosine of [latex]\frac{\pi }{3}[/latex].

Figure 9.  The sine of [latex]\frac{\pi }{3}[/latex] equals the cosine of [latex]\frac{\pi }{6}[/latex] and vice versa.

This result should not be surprising because, as we see from Figure 9, the side opposite the angle of [latex]\frac{\pi }{3}[/latex] is also the side adjacent to [latex]\frac{\pi }{6}[/latex], so [latex]\sin \left(\frac{\pi }{3}\right)[/latex] and [latex]\cos \left(\frac{\pi }{6}\right)[/latex] are exactly the same ratio of the same two sides, [latex]\sqrt{3}s[/latex] and [latex]2s[/latex]. Similarly, [latex]\cos \left(\frac{\pi }{3}\right)[/latex] and [latex]\sin \left(\frac{\pi }{6}\right)[/latex] are also the same ratio using the same two sides, [latex]s[/latex] and [latex]2s[/latex].

The interrelationship between the sines and cosines of [latex]\frac{\pi }{6}[/latex] and [latex]\frac{\pi }{3}[/latex] also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to [latex]\pi [/latex], and the right angle is [latex]\frac{\pi }{2}[/latex], the remaining two angles must also add up to [latex]\frac{\pi }{2}[/latex]. That means that a right triangle can be formed with any two angles that add to [latex]\frac{\pi }{2}[/latex] —in other words, any two complementary angles. So we may state a cofunction identity : If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure 10.

Figure 10.  Cofunction identity of sine and cosine of complementary angles

Using this identity, we can state without calculating, for instance, that the sine of [latex]\frac{\pi }{12}[/latex] equals the cosine of [latex]\frac{5\pi }{12}[/latex], and that the sine of [latex]\frac{5\pi }{12}[/latex] equals the cosine of [latex]\frac{\pi }{12}[/latex]. We can also state that if, for a certain angle [latex]t[/latex], [latex]\cos \text{ }t=\frac{5}{13}[/latex], then [latex]\sin \left(\frac{\pi }{2}-t\right)=\frac{5}{13}[/latex] as well.

A General Note: Cofunction Identities

The cofunction identities in radians are listed in the table below.

How To: Given the sine and cosine of an angle, find the sine or cosine of its complement.

• To find the sine of the complementary angle, find the cosine of the original angle.
• To find the cosine of the complementary angle, find the sine of the original angle.

Example 5: Using Cofunction Identities

Write the following as an equivalent cosine expression: [latex]\sin \left(\frac{5\pi}{12}\right)[/latex].

According to the cofunction identities for sine and cosine,

[latex]\sin t=\cos \left(\frac{\pi }{2}-t\right)[/latex].

[latex]\cos \left(\frac{\pi }{2}-\frac{5\pi}{12}\right)=\cos\left(\frac{\pi}{12}\right)[/latex].

Example 6: Using Cofunction Identities

Write the following as an equivalent tangent expression: [latex]\cot\left(25^\circ\right)[/latex].

According to the cofunction identities for tangent and cotangent,

[latex]\tan t=\cot \left(90^\circ-t\right)[/latex].

[latex]\tan \left(90^\circ-25^\circ\right)=\tan\left(65^\circ\right)[/latex].

Write the following as an equivalent cosecant expression: [latex]\sec\left(68^\circ\right)[/latex].

[latex]\csc\left(22^\circ\right)[/latex]

Using Trigonometric Functions

In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.

How To: Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides.

• For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.
• Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
• Using the value of the trigonometric function and the known side length, solve for the missing side length.

Example 7: Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in Figure 11.

We know the angle and the opposite side, so we can use the tangent to find the adjacent side.

[latex]\tan \left(30^\circ \right)=\frac{7}{a}[/latex]

We rearrange to solve for [latex]a[/latex].

[latex]\begin{align}a&=\frac{7}{\tan \left(30^\circ \right)} \\ &=12.1\end{align}[/latex]

We can use the sine to find the hypotenuse.

[latex]\sin \left(30^\circ \right)=\frac{7}{c}[/latex]

Again, we rearrange to solve for [latex]c[/latex].

[latex]\begin{align}c&=\frac{7}{\sin \left(30^\circ \right)} \\ &=14 \end{align}[/latex]

A right triangle has one angle of [latex]\frac{\pi }{3}[/latex] and a hypotenuse of 20. Find the unknown sides and angle of the triangle.

[latex]\text{adjacent}=10[/latex]; [latex]\text{opposite}=10\sqrt{3}[/latex] ; missing angle is [latex]\frac{\pi }{6}[/latex]

Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer’s eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer’s eye.  In the figure below, [latex]\alpha[/latex] represents the angle of elevation and [latex]\beta[/latex] represents the angle of depression .

How To: Given a tall object, measure its height indirectly.

• Make a sketch of the problem situation to keep track of known and unknown information.
• Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.
• At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.
• Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.
• Solve the equation for the unknown height.

Example 8: Measuring a Distance Indirectly

To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures the angle of elevation to be [latex]57^\circ [/latex], as shown in Figure 13. Find the height of the tree.

We know that the angle of elevation is [latex]57^\circ [/latex] and the adjacent side is 30 ft long. The opposite side is the unknown height.

The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of [latex]57^\circ [/latex], letting [latex]h[/latex] be the unknown height.

[latex]\begin{align}&\tan \theta =\frac{\text{opposite}}{\text{adjacent}} \\ &\tan\left(57^\circ \right)=\frac{h}{30}&& \text{Solve for }h. \\ &h=30\tan \left(57^\circ \right)&& \text{Multiply}.\\ &h\approx 46.2&& \text{Use a calculator}. \end{align}[/latex]

The tree is approximately 46 feet tall.

How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building with an angle of elevation [latex]\frac{5\pi }{12}[/latex] ? Round to the nearest foot.

About 52 ft

Inverse Trigonometric Function on a calculator

If two sides of a right triangle are given, an inverse trigonometric function can be used to find an acute angle in a triangle. There is an inverse trig button on a calculator that appears above the normal sin, cos, tan buttons on a calculator. A shift or second key is required to select these, and they are [latex]\sin^{-1},\cos^{-1},\tan^{-1}[/latex]. The calculator will return an angle in either radians or degrees, depending on what mode your calculator is in.  We will need the inverse tangent function for bearing.

Now we will will look at some application problems that involve bearings and right triangles.

Example 9: Find the bearing

A semi travels east for 8 miles, makes a right turn, and then travels south for 11 more miles. What is the bearing from the starting point to the semi’s current location?

First we need to draw a picture of the problem:

We want to find the angle [latex]\alpha[/latex] that is inside the triangle. To do this, we will set up a trig definition that relates the two given sides. We will choose tangent. Then we will take the inverse tangent to find the angle: [latex]\begin{align} &\tan\left(\alpha\right)=\frac{11}{8} \\& \tan^{-1}\left(\tan\left(\alpha\right)\right)=\tan^{-1}\left(\frac{11}{8}\right) \\ &\alpha=\tan^{-1}\left(\frac{11}{8}\right) \\& \alpha\approx 54^\circ \end{align}[/latex]

The bearing must be measured from south since the triangle is drawn in the fourth quadrant. On the picture, we must find [latex]\beta[/latex]. Since [latex]\alpha+\beta=90^\circ[/latex], then [latex]\beta=90^\circ-\alpha=36^\circ[/latex]. Therefore our bearing is [latex]S 36^\circ E[/latex].

A cyclist travels west for 7 miles, makes a right turn and travels north for 4 miles. What is the bearing from the starting point to the cyclist’s current position? Round your angle to one decimal place.

[latex]N 60.3^\circ W[/latex]

Example 10: Find the bearing

A jeep travels on a bearing of [latex]N 40^\circ W[/latex] for 6 miles. How far north and how far west is the jeep from its starting point? Round your answers to one decimal place.

We need to find the angle [latex]\alpha[/latex] inside of the triangle before we set up a trig equation: [latex]\alpha=90^\circ-40^\circ=50^\circ[/latex].

Next we will set up a trig function to find the indicated sides of the triangle:

First we will find the number of miles traveled North:

[latex]\begin{align} & \sin\left(50^\circ\right)=\frac{N}{6} \\& 6\sin\left(50^\circ\right)=N \\ & N\approx 4.6\text{ miles} \end{align}[/latex]

Now we will find the number of miles traveled west:

[latex]\begin{align} & \cos\left(50^\circ\right)=\frac{W}{6} \\& 6\cos\left(50^\circ\right)=W \\ & W\approx 3.9\text{ miles} \end{align}[/latex]

A jogger travels on a bearing of [latex]S 53.13^\circ E[/latex] for 5 miles. How far east and how far south is the jogger from their starting point? Round your answers to the nearest mile.

4 miles east, 3 miles south

Convert Percent Grade Into Degrees

Percent grade is often seen on roadways or trails, and it is a measure of steepness.  For example, if a road has a 6% grade, this means the road rises 6 feet over a horizontal distance (run) of 100 feet.  Percent grade is found by dividing the rise over the run as shown in Figure 14. The rise and run are also the opposite and adjacent sides.  We can find [latex]\theta[/latex] by taking the inverse tangent.

In order to calculate the degree measurement of a percent grade, first change the percent into a decimal by dividing by 100. Then take the inverse tangent of this decimal, and this will give the angle in degrees.  This is actually the angle of elevation , which we studied earlier in this section!

How To: Convert Percent Grade inTo Degrees

Make sure your calculator is in degree mode. Given the percent grade, the angle [latex]\theta[/latex] can be calculated using:

[latex]\theta=\tan^{-1}\left(\dfrac{\text{percent grade}}{100}\right)[/latex]

Example 11: Percent Grade to Degrees

Filbert Street, the steepest street in San Francisco, has a 31.5% grade.  What is the angle the street makes with a horizontal line, rounded to one decimal place?

[latex]\theta=\tan^{-1}\left(\dfrac{31.5}{100}\right)\approx 17.5^\circ[/latex]

A road has a grade of 10.9%.  What is the angle the road makes with a horizontal line, rounded to one decimal place?

[latex]\text{About }6.2^\circ[/latex]

Key Equations

Key concepts.

• We can define trigonometric functions as ratios of the side lengths of a right triangle.
• The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle.
• We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur.
• Any two complementary angles could be the two acute angles of a right triangle.
• If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa.
• We can use trigonometric functions of an angle to find unknown side lengths.
• Select the trigonometric function representing the ratio of the unknown side to the known side.
• Right-triangle trigonometry permits the measurement of inaccessible heights and distances.
• The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known.
• Bearings are measured from either from N or S, depending on the first letter of the bearing.
• Percent grade can be expressed as a degree, which is the angle of elevation.

Section 4.3 Homework Exercises

2. When a right triangle with a hypotenuse of 1 is placed in the unit circle, which sides of the triangle correspond to the x- and y-coordinates?

3. The tangent of an angle compares which sides of the right triangle?

4. What is the relationship between the two acute angles in a right triangle?

5. Explain the cofunction identity.

For the following exercises, evaluate the expression.  Rationalize the denominator if necessary.

6. [latex]\sin\frac{\pi}{4}\cos\frac{\pi}{4}-\tan\frac{\pi}{3}[/latex]

7. [latex]\sin\frac{\pi}{4}\cos\frac{\pi}{3}-\tan\frac{\pi}{4}[/latex]

8. [latex]2\sec^{2}\frac{\pi}{4}+\cot^{2}\frac{\pi}{3}[/latex]

9. [latex]3\csc^{2}\frac{\pi}{3}+\cot^{2}\frac{\pi}{4}[/latex]

For the following exercises, use cofunctions of complementary angles.

10. [latex]\cos \left(34^\circ\right)=\sin \left(\text{__}^\circ\right)[/latex]

11. [latex]\cos \left(\frac{\pi }{3}\right)=\sin \text{(___)}[/latex]

12. [latex]\csc \left(21^\circ\right)=\sec \left(\text{___}^\circ \right)[/latex]

13. [latex]\tan \left(\frac{\pi }{4}\right)=\cot \left(\text{__}\right)[/latex]

For the following exercises, find the lengths of the missing sides if side [latex]a[/latex] is opposite angle [latex]A[/latex], side [latex]b[/latex] is opposite angle [latex]B[/latex], and side [latex]c[/latex] is the hypotenuse.

14. [latex]\cos B=\frac{4}{5},a=10[/latex]

15. [latex]\sin B=\frac{1}{2}, a=20[/latex]

16. [latex]\tan A=\frac{5}{12},b=6[/latex]

17. [latex]\tan A=100,b=100[/latex]

18. [latex]\sin B=\frac{1}{\sqrt{3}}, a=2[/latex]

19. [latex]a=5,\measuredangle A={60}^{\circ }[/latex]

20. [latex]c=12,\measuredangle A={45}^{\circ }[/latex]

For the following exercises, use Figure 14 to evaluate each trigonometric function of angle [latex]A[/latex].

21. [latex]\sin A[/latex]

22. [latex]\cos A[/latex]

23. [latex]\tan A[/latex]

24. [latex]\csc A[/latex]

25. [latex]\sec A[/latex]

26. [latex]\cot A[/latex]

For the following exercises, use Figure 15 to evaluate each trigonometric function of angle [latex]A[/latex].

27. [latex]\sin A[/latex]

28. [latex]\cos A[/latex]

29. [latex]\tan A[/latex]

30. [latex]\csc A[/latex]

31. [latex]\sec A[/latex]

32. [latex]\cot A[/latex]

For the following exercises, solve for the unknown sides of the given triangle.

For the following exercises, use a calculator to find the length of each side to four decimal places.

41. [latex]b=15,\measuredangle B={15}^{\circ }[/latex]

42. [latex]c=200,\measuredangle B={5}^{\circ }[/latex]

43. [latex]c=50,\measuredangle B={21}^{\circ }[/latex]

44. [latex]a=30,\measuredangle A={27}^{\circ }[/latex]

45. [latex]b=3.5,\measuredangle A={78}^{\circ }[/latex]

50. A radio tower is located 400 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is [latex]36^\circ [/latex], and that the angle of depression to the bottom of the tower is [latex]23^\circ [/latex]. How tall is the tower?

51. A radio tower is located 325 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is [latex]43^\circ [/latex], and that the angle of depression to the bottom of the tower is [latex]31^\circ [/latex]. How tall is the tower?

52. A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is [latex]15^\circ [/latex], and that the angle of depression to the bottom of the tower is [latex]2^\circ [/latex]. How far is the person from the monument?

53. A 400-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is [latex]18^\circ [/latex], and that the angle of depression to the bottom of the tower is [latex]3^\circ [/latex]. How far is the person from the monument?

54. There is an antenna on the top of a building. From a location 300 feet from the base of the building, the angle of elevation to the top of the building is measured to be [latex]40^\circ [/latex]. From the same location, the angle of elevation to the top of the antenna is measured to be [latex]43^\circ [/latex]. Find the height of the antenna.

55. There is lightning rod on the top of a building. From a location 500 feet from the base of the building, the angle of elevation to the top of the building is measured to be [latex]36^\circ [/latex]. From the same location, the angle of elevation to the top of the lightning rod is measured to be [latex]38^\circ [/latex]. Find the height of the lightning rod.

56. A 33-ft ladder leans against a building so that the angle between the ground and the ladder is [latex]80^\circ [/latex]. How high does the ladder reach up the side of the building?

57. A 23-ft ladder leans against a building so that the angle between the ground and the ladder is [latex]80^\circ [/latex]. How high does the ladder reach up the side of the building?

58. The angle of elevation to the top of a building in New York is found to be 9 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building.

59. The angle of elevation to the top of a building in Seattle is found to be 2 degrees from the ground at a distance of 2 miles from the base of the building. Using this information, find the height of the building.

60. Assuming that a 370-foot tall giant redwood grows vertically, if I walk a certain distance from the tree and measure the angle of elevation to the top of the tree to be [latex]60^\circ [/latex], how far from the base of the tree am I?

61. A car travels west for 5 miles, turns left, and then travels south for 9 miles.  What is the bearing from car’s starting position to its current position?  Round your answer to two decimal places.

62. A truck travels east for 4 miles, turns left, and then travels north for 6 miles.  What is the bearing from the truck’s starting position to its current position? Round your answer to two decimal places.

63. An ant travels on a bearing of [latex]N 22^\circ E[/latex] for 36 inches.  How far east and how far north is the ant from its starting position?  Round your answers to two decimal places.

64. A spider crawls on a bearing of [latex]S 34^\circ W[/latex] for 20 inches.  How far west and how far south is the spider from its starting point? Round your answers to two decimal places.

65. Originally built in 1901 by Colonel J.W. Eddy, Angels Flight in Los Angeles is said to be the world’s shortest incorporated railway.  The counterbalanced cars, controlled by cables, travel a 33% grade for 315 feet.  What is the angle the track makes with a horizontal line, rounded to one decimal place?

66. The Saluda Grade is the steepest standard-gauge mainline railway grade in the United States.  Between Melrose and Saluda, North Carolina, the maximum grade is 4.9% for about 300 feet.  What is the angle the track makes with a horizontal line, rounded to one decimal place?

• Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution
• Introduction to Trigonometric Functions Using Triangles. Authored by : Mathispower4u. Located at : https://youtu.be/Ujyl_zQw2zE . License : All Rights Reserved . License Terms : Standard YouTube License
• Cofunction Identities. Authored by : Mathispower4u. Located at : https://youtu.be/_gkuml--4_Q . License : All Rights Reserved . License Terms : Standard YouTube License
• Example: Determine the Length of a Side of a Right Triangle Using a Trig Equation. Authored by : Mathispower4u. Located at : https://youtu.be/8jU2R3BuR5E . License : All Rights Reserved . License Terms : Standard YouTube License

Right Triangle Trigonometry Calculator

Basics of trigonometry, right triangles trigonometry calculations, example of right triangle trigonometry calculations with steps, more trigonometry and right triangles calculators (and not only).

The right triangle trigonometry calculator can help you with problems where angles and triangles meet: keep reading to find out:

• The basics of trigonometry;
• How to calculate a right triangle with trigonometry;
• A worked example of how to use trigonometry to calculate a right triangle with steps;

And much more!

Trigonometry is a branch of mathematics that relates angles to the length of specific segments . We identify multiple trigonometric functions: sine, cosine, and tangent, for example. They all take an angle as their argument, returning the measure of a length associated with the angle itself. Using a trigonometric circle , we can identify some of the trigonometric functions and their relationship with angles.

As you can see from the picture, sine and cosine equal the projection of the radius on the axis, while the tangent lies outside the circle. If you look closely, you can identify a right triangle using the elements we introduced above: let's discover the relationship between trigonometric functions and this shape.

Consider an acute angle in the trigonometric circle above: notice how you can build a right triangle where:

• The radius is the hypotenuse; and
• The sine and cosine are the catheti of the triangle.

α \alpha α is one of the acute angles, while the right angle lies at the intersection of the catheti (sine and cosine)

Let this sink in for a moment: the length of the cathetus opposite from the angle α \alpha α is its sine , sin ⁡ ( α ) \sin(\alpha) sin ( α ) ! You just found an easy and quick way to calculate the angles and sides of a right triangle using trigonometry.

The complete relationships between angles and sides of a right triangle need to contain a scaling factor, usually the radius (the hypotenuse). Identify the opposite and adjacent . We can then write:

By switching the roles of the legs, you can find the values of the trigonometric functions for the other angle.

Taking the inverse of the trigonometric functions , you can find the values of the acute angles in any right triangle.

Using the three equations above and a combination of sides, angles, or other quantities, you can solve any right triangle . The cases we implemented in our calculator are:

• Solving the triangle knowing two sides ;
• Solving the triangle knowing one angle and one side ; and
• Solving the triangle knowing the area and one side .

Take a right triangle with hypotenuse c = 5 c = 5 c = 5 and an angle α = 38 ° \alpha=38\degree α = 38° . Surprisingly enough, this is enough data to fully solve the right triangle! Follow these steps:

• Calculate the third angle: β = 90 ° − α \beta = 90\degree - \alpha β = 90° − α .
• sin ⁡ ( α ) = 0.61567 \sin(\alpha) = 0.61567 sin ( α ) = 0.61567 .
• o p p o s i t e = sin ⁡ ( α ) ⋅ h y p o t e n u s e = 0.61567 ⋅ 5 = 3.078 \mathrm{opposite} = \sin(\alpha)\cdot\mathrm{hypotenuse} = 0.61567 \cdot 5 = 3.078 opposite = sin ( α ) ⋅ hypotenuse = 0.61567 ⋅ 5 = 3.078 .
• a d j a c e n t = 0.788 ⋅ 5 = 3.94 \mathrm{adjacent} = 0.788\cdot 5 = 3.94 adjacent = 0.788 ⋅ 5 = 3.94 .

If you liked our right triangle trigonometry calculator, why not try our other related tools? Here they are:

• The trigonometry calculator ;
• The cosine triangle calculator ;
• The sine triangle calculator ;
• The trig triangle calculator ;
• The trig calculator ;
• The sine cosine tangent calculator ;
• The tangent ratio calculator ; and
• The tangent angle calculator .

How do I apply trigonometry to a right triangle?

To apply trigonometry to a right triangle, remember that sine and cosine correspond to the legs of a right triangle . To solve a right triangle using trigonometry:

• sin(α) = opposite/hypotenuse ; and
• cos(α) = adjacent/hypotenuse .
• By taking the inverse trigonometric functions , we can find the value of the angle α .
• You can repeat the procedure for the other angle.

What is the hypotenuse of a triangle with α = 30° and opposite leg a = 3?

The length of the hypotenuse is 6 . To find this result:

• Calculate the sine of α : sin(α) = sin(30°) = 1/2 .
• Apply the following formula: sin(α) = opposite/hypotenuse hypotenuse = opposite/sin(α) = 3 · 2 = 6 .

Can I apply right-triangle trigonometric rules in a non-right triangle?

Not directly: to apply the relationships between trigonometric functions and sides of a triangle, divide the shape alongside one of the heights lying inside it. This way, you can split the triangle into two right triangles and, with the right combination of data, solve it!

Divide complex numbers

Equation of a sphere, grams to cups.

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Chapter 2: Trigonometric Ratios

Exercises: 2.2 Right Triangle Trigonometry

Exercises homework 2.2.

• Use measurements to calculate the trigonometric ratios for acute angles #1-10, 57-60
• Use trigonometric ratios to find unknown sides of right triangles #11-26
• Solve problems using trigonometric ratios #27-34, 41-46
• Use trig ratios to write equations relating the sides of a right triangle #35-40
• Use relationships among the trigonometric ratios #47-56, 61-68

Suggested Homework Problems

Here are two right triangles with a [latex]65 °[/latex] angle.

• Measure the sides [latex]AB[/latex] and [latex]BC[/latex] with a ruler. Use the lengths to estimate [latex]\sin 65°{.}[/latex]
• Measure the sides [latex]AD[/latex] and [latex]DE[/latex] with a ruler. Use the lengths to estimate [latex]\sin 65°{.}[/latex]
• Use your calculator to look up [latex]\sin 65°{.}[/latex] Compare your answers. How close were your estimates?

Use the figure in Problem 1 to calculate two estimates each for the cosine and tangent of [latex]65 °{.}[/latex] Compare your estimates to your calculator’s values for [latex]\cos 65°[/latex] and [latex]\tan 65°{.}[/latex]

Here are two right triangles with a [latex]40 °[/latex] angle.

• Measure the sides [latex]AB[/latex] and [latex]AC[/latex] with a ruler. Use the lengths to estimate [latex]\cos 40°{.}[/latex]
• Measure the sides [latex]AD[/latex] and [latex]AE[/latex] with a ruler. Use the lengths to estimate [latex]\cos 40°{.}[/latex]
• Use your calculator to look up [latex]\cos 40°{.}[/latex] Compare your answers. How close were your estimates?

Use the figure in Problem 2 to calculate two estimates each for the cosine and tangent of [latex]40 °{.}[/latex] Compare your estimates to your calculator’s values for [latex]\sin 40°[/latex] and [latex]\tan 40°{.}[/latex]

Exercise Group

For the right triangles in Problems 5–10,

• Find the length of the unknown side.
• Find the sine, cosine, and tangent of [latex]\theta \text{.}[/latex] Round your answers to four decimal places.

For Problems 11–16,

• Sketch and label the sides of a right triangle with angle [latex]\theta\text{.}[/latex]
• Sketch and label another right triangle with angle [latex]\theta[/latex] and longer sides.

[latex]\cos \theta = \dfrac{3}{5}[/latex]

[latex]\tan \theta = \dfrac{7}{2}[/latex]

[latex]\tan \theta = \dfrac{11}{4}[/latex]

[latex]\sin \theta = \dfrac{4}{9}[/latex]

[latex]\sin \theta = \dfrac{1}{9}[/latex]

[latex]\cos \theta = \dfrac{7}{8}[/latex]

For Problems 17–22, use one of the three trigonometric ratios to find the unknown side of the triangle. Round your answer to hundredths.

For Problems 23–26, sketch and label a right triangle with the given properties.

One angle is [latex]40°{,}[/latex] the side opposite that angle is 8 inches

One angle is [latex]65°{,}[/latex] the side adjacent to that angle is 30 yards

One angle is [latex]28°{,}[/latex] the hypotenuse is 56 feet

One leg is [latex]15[/latex] meters, the hypotenuse is [latex]18[/latex] meters

For Problems 27–34,

• Sketch a right triangle that illustrates the situation. Label your sketch with the given information.
• Choose the appropriate trig ratio and write an equation, then solve the problem.

To measure the height of cloud cover, airport controllers fix a searchlight to shine a vertical beam on the clouds. The searchlight is [latex]120[/latex] yards from the office. A technician in the office measures the angle of elevation to the light on the cloud cover at [latex]54.8°{.}[/latex] What is the height of the cloud cover?

To measure the distance across a canyon, Evel first sights an interesting rock directly opposite on the other side. He then walks [latex]200[/latex] yards down the rim of the canyon and sights the rock again, this time at an angle of [latex]18.5°[/latex] from the canyon rim. What is the width of the canyon?

A salvage ship is searching for the wreck of a pirate vessel on the ocean floor. Using sonar, they locate the wreck at an angle of depression of [latex]36.2°{.}[/latex] The depth of the ocean at their location is [latex]260[/latex] feet. How far should they move so that they are directly above the wrecked vessel?

Ramps for wheelchairs should be no steeper than an angle of [latex]6°{.}[/latex] How much horizontal distance should be allowed for a ramp that rises [latex]5[/latex] feet in height?

The radio signal from a weather balloon indicates that it is [latex]1500[/latex] meters from a meteorologist on the ground. The angle of elevation to the balloon is [latex]48°{.}[/latex] What is the balloon’s altitude?

According to Chinese legend, around 200 BC, the general Han Xin used a kite to determine the distance from his location to an enemy palace. He then dug a secret tunnel which emerged inside the palace. When the kite was directly above the palace, its angle of elevation was [latex]27°[/latex] and the string to the kite was [latex]1850[/latex] feet long. How far did Han Xin’s troops have to dig?

A cable car on a ski lift traverses a horizontal distance of [latex]1800[/latex] meters at an angle of [latex]38°{.}[/latex] How long is the cable?

Zelda is building the loft on her summer cottage. At its central point, the height of the loft is [latex]8[/latex] feet, and the pitch of the roof should be [latex]24°{.}[/latex] How long should the rafters be?

For Problems 35–40, use a trig ratio to write an equation for [latex]x[/latex] in terms of [latex]\theta{.}[/latex]

For Problems 41–44, find the altitude of the triangle. Round your answer to two decimal places.

For Problems 45 and 46, find the length of the chord [latex]AB{.}[/latex] Round your answer to two decimal places.

For Problems 47–50, fill in the table.

• In each of the figures for Problems 47-50, what is the relationship between the angles [latex]\theta[/latex] and [latex]\phi{?}[/latex]
• Study the tables for Problems 47-50. What do you notice about the values of sine and cosine for the angles [latex]\theta[/latex] and [latex]\phi{?}[/latex] Explain why this is true.

There is a relationship between the tangent, the sine, and the cosine of any angle. Study the tables for Problems 47-50 to discover this relationship. Write your answer as an equation.

• Use the figure to explain what happens to [latex]\tan \theta[/latex] as [latex]\theta[/latex] increases, and why.
• Use the figure to explain what happens to [latex]\cos \theta[/latex] as [latex]\theta[/latex] increases, and why.
• What happens to [latex]\tan \theta[/latex] as [latex]\theta[/latex] increases?
• What value does your calculator give for [latex]\tan 90°{?}[/latex] Why?

Explain why it makes sense that [latex]\sin 0° = 0[/latex] and [latex]\sin 90° = 1{.}[/latex] Use a figure to illustrate your explanation.

Explain why it makes sense that [latex]\cos 0° = 1[/latex] and [latex]\cos 90° = 0{.}[/latex] Use a figure to illustrate your explanation

For Problems 57–60, explain why the trigonometric ratio is not correct.

[latex]\sin \theta = \dfrac{5}{9}[/latex]

[latex]\tan \theta = \dfrac{4}{7}[/latex]

[latex]\cos \theta = \dfrac{21}{20}[/latex]

[latex]\sin \theta = \dfrac{8}{10}[/latex]

For Problems 61–64, sketch and label a right triangle, then fill in the blank.

If [latex]\sin \theta = 0.2358\text{,}[/latex]  then [latex]\cos (90^{o} - \theta)=\underline{\qquad} \text{,}[/latex]

• If [latex]\cos \alpha = \dfrac{3}{11} \text{,}[/latex] then [latex]\underline{\qquad} (90° - \alpha) = \dfrac{3}{11}\text{.}[/latex]
• If [latex]\sin 42° = n\text{,}[/latex] then [latex]\cos \underline{\qquad} = n\text{.}[/latex]
• If [latex]\cos 13° = z\text{,}[/latex] then [latex]\sin \underline{\qquad} = z\text{.}[/latex]
• If [latex]\cos \beta = \dfrac{2}{\sqrt{7}}{,}[/latex] then [latex]\sin (90° - \beta) =\underline{\qquad} {.}[/latex]
• If [latex]\sin \phi = 0.693{,}[/latex] then [latex](90° - \phi) = 0.693{.}[/latex]
• If [latex]\cos 87° = p{,}[/latex] then [latex]\sin \underline{\qquad} = p{.}[/latex]
• If [latex]\sin 59° =w{,}[/latex] then [latex]\cos \underline{\qquad} = w{.}[/latex]
• If [latex]\sin \phi = \dfrac{5}{13}[/latex] and [latex]\cos \phi = \dfrac{12}{13}{,}[/latex] then [latex]\tan \phi =\underline{\qquad} {.}[/latex]
• If [latex]\cos \beta = \dfrac{1}{\sqrt{10}}{,}[/latex] and [latex]\sin \beta = \dfrac{3}{\sqrt{10}}{,}[/latex] then [latex]\tan \beta = \underline{\qquad}{.}[/latex]
• If [latex]\tan B = \dfrac{2}{\sqrt{5}}[/latex] and [latex]\cos B = \dfrac{\sqrt{5}}{3}{,}[/latex] then [latex]\sin B =\underline{\qquad}{.}[/latex]
• If [latex]\sin W = \sqrt{\dfrac{3}{7}}[/latex] and [latex]\tan W = \dfrac{\sqrt{3}}{2}{,}[/latex] then [latex]\cos W =\underline{\qquad}{.}[/latex]
• If [latex]\cos \theta = \dfrac{2}{\sqrt{10}}[/latex] and [latex]\sin \theta = \sqrt{\dfrac{3}{5}}{,}[/latex] then [latex]\tan \theta = \underline{\qquad}{.}[/latex]
• If [latex]\sin \alpha = \dfrac{\sqrt{2}}{4}{,}[/latex] and [latex]\cos \alpha =\dfrac{\sqrt{14}}{4}{,}[/latex] then [latex]\tan \alpha = \underline{\qquad}{.}[/latex]
• If [latex]\tan A = \dfrac{\sqrt{7}}{3}[/latex] and [latex]\cos A = \dfrac{3}{4}{,}[/latex] then [latex]\sin A =\underline{\qquad}{.}[/latex]
• If [latex]\sin V = \sqrt{\dfrac{10}{5}}[/latex] and [latex]\tan V = \dfrac{2}{5}{,}[/latex] then [latex]\cos V =\underline{\qquad}{.}[/latex]

Explain why the cosine of a [latex]73°[/latex] angle is always the same, no matter what size triangle the angle is in. Illustrate your explanation with a sketch.

• If you plotted the points in your table, would they lie on a straight line? Why or why not?
• What is the slope of the line through the origin and point [latex]P{?}[/latex]
• What is the tangent of the angle [latex]\theta{?}[/latex]
• On the same grid, sketch an angle whose tangent is [latex]\dfrac{8}{5}.[/latex]
• Use your calculator to complete the table. Round your answers to hundredths.
• Use the values of tan [latex]\theta[/latex] to sketch all the angles listed in the table. Locate the vertex of each angle at the origin and the initial side along the positive [latex]x[/latex]-axis.

Trigonometry Copyright © 2024 by Bimal Kunwor; Donna Densmore; Jared Eusea; and Yi Zhen. All Rights Reserved.

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11.1: The Trigonometric Ratios

• Last updated
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• Page ID 122907

• Katherine Yoshiwara
• Los Angeles Pierce College

2.1 Side and Angle Relationships

Homework 2.1.

1. The sum of the angles is not \(180^{\circ}\)

3. The exterior angle is not equal to the sum of the opposite interior angles.

5. The sum of the acute angles is not \(90^{\circ}\)

7. The largest side is not opposite the largest angle.

9. The Pythagorean theorem is not satisfied.

11. \(5^2+12^2=13^2\), but the angle opposite the side of length 13 is \(85^{\circ}\).

13. \(4 < x < 16\)

15. \(0 < x < 16\)

19. \(6\sqrt{2}\)in

21. \(w = 6\sqrt{10}\)in

25. \(\sqrt{3}\)

33. The distance from \((0,0)\) to \((3,3)\) is \(3 \sqrt{2}\), and the distance from \((3,3)\) to \((6,0)\) is also \(3 \sqrt{2}\), so the triangle is isosceles. The distance from \((0,0)\) to \((6,0)\) is 6 , and \((3 \sqrt{2})^2+(3 \sqrt{2})^2=6^2\) so the triangle is a right triangle.

37. \(\alpha=30^{\circ}, \beta=60^{\circ}, h=\sqrt{3}\)

39. \(8 \sqrt{3}\) in

a \((-1,0)\) and \((1,0) ; 2\)

b \(\sqrt{(p+1)^2+q^2}\) and \(\sqrt{(p-1)^2+q^2}\)

\begin{aligned} \left(\sqrt{(p+1)^2+q^2}\right)^2 & +\left(\sqrt{(p-1)^2+q^2}\right)^2 \\ & =p^2+2 p+1+q^2+p^2-2 p+1+q^2 \\ & =2 p^2+2+2 q^2=2+2\left(p^2+q^2\right) \\ & =2+2(1)=4 \end{aligned}

2.2 Right Triangle Trigonometry

Homework 2.2.

a \(4\sqrt{13} \approx 14.42\)

b \(\sin \theta = 0.5547, \cos \theta = 0.8321, \tan \theta = 0.6667

a \(4\sqrt{15} \approx 15.49\)

b \(\sin \theta = 0.9682, \cos \theta = 0.2500, \tan \theta = 3.8730\)

a \(2\sqrt{67} \approx 15.49\)

b \(\sin \theta = 0.2116, \cos \theta = 0.9774, \tan \theta = 0.2165\)

(Answers may vary)

b \(\tan 54.8^{\circ} = \dfrac{h}{20}, 170.1\) yd

b \(\tan 36.2^{\circ} = \dfrac{260}{d}, 355.2\) ft

b \(\sin 48^{\circ} = \dfrac{a}{1500}, 1114.7\)m

b \(\cos 38^{\circ} = \dfrac{1800}{x}, 2284.2\)m

35. \(x = \dfrac{82}{\tan \theta}\)

37. \(x = 11 \sin \theta\)

39. \(x = \dfrac{9}{\cos \theta}\)

41. \(36 \sin 25^{\circ} \approx 15.21\)

43. \(46 \sin 20^{\circ} \approx 15.73\)

45. \(12 \sin 40^{\circ} \approx 7.71\)

a \(\theta\) and \(\phi\) are complements.

b \(\sin \theta=\cos \phi\) and \(\cos \theta=\sin \phi\). The side opposite \(\theta\) is the side adjacent to \(\phi\), and vice versa.

a As \(\theta\) increases, \(\tan \theta\) increases also. The side opposite \(\theta\) increases in length while the side adjacent to \(\theta\) remains fixed.

b As \(\theta\) increases, \(\cos \theta\) decreases. The side adjacent to \(\theta\) remains fixed while the hypotenuse increases in length.

55. As \(\theta\) decreases toward \(0^{\circ}\), the side opposite \(\theta\) approaches a length of 0, so \(\sin \theta\) approaches 0. But as \(\theta\) increases toward \(90^{\circ}\), the length of the side opposite \(\theta\) approaches the length of the hypotenuse, so \(\sin \theta\) approaches 1.

57. The triangle is not a right triangle.

59. \(\dfrac{21}{20}\) is the ratio of hypotenuse to the adjacent side, which is the reciprocal of \(\cos \theta\).

a \(0.2358\)

c \(48^{\circ}\)

d \(77^{\circ}\)

a \(\dfrac{5}{12}\)

c \(\dfrac{2}{3}\)

d \(\dfrac{2}{\sqrt{7}}\)

65. Although the triangles may differ in size, the ratio of the side adjacent to the angle to the hypotenuse of the triangle remains the same because the triangles would all be similar, and hence corresponding sides are proportional.

a \(\dfrac{2}{3}\)

b \(\dfrac{2}{3}\)

2.3 Solving Right Triangles

Homework 2.3.

1. \(A = 61^{\circ}, a = 25.26, c = 28.88\)

3. \(A = 68^{\circ}, a = 0.93, b = 0.37\)

b \(B = 48^{\circ}, a = 17.4, b = 19.3\)

b \(A=57^{\circ}, b=194.4, c=357.7\)

b \(B=78^{\circ}, b=18.8, c=19.2\)

• Solve \(\sin 53.7^{\circ} = \dfrac{8.2}{c}\) for \(c\).
• Solve \(\tan 53.7^{\circ} = \dfrac{8.2}{a}\) for \(a\).
• Subtract \(53.7^{\circ}\) from \(90^{\circ}\) to find \(A\).

• Solve \(\cos 25^{\circ} = \dfrac{40}{c}\) for \(c\).
• Solve \(\tan 25^{\circ} = \dfrac{a}{40}\) for \(a\).
• Subtract \(25^{\circ}\) from \(90^{\circ}\) to find \(B\).

• Solve \(\sin 64.5^{\circ} = \dfrac{a}{24}\) for \(a\).
• Solve \(\cos 64.5^{\circ} = \dfrac{b}{24}\) for \(b\).
• Subtract \(64.5^{\circ}\) from \(90^{\circ}\) to find \(B\).

17. \(74.2^{\circ}\)

19. \(56.4^{\circ}\)

21. \(66.0^{\circ}\)

23. \(11.5^{\circ}\)

25. \(56.3^{\circ}\)

27. \(73.5^{\circ}\)

29. \(\cos 15^{\circ} = 0.9659\) and \(\cos ^{-1} 0.9659 = 15^{\circ}\)

31. \(\tan 65^{\circ} = 2.1445\) and \(\tan ^{-1} 2.1445 = 65^{\circ}\)

33. \(\sin ^{-1} (0.6) \approx 36.87^{\circ}\) is the angle whose sine is 0.6. \((\sin 6^{\circ})^{-1} \approx 9.5668\) is the reciprocal of \(\sin 6^{\circ}\).

b \(\sin \theta = \dfrac{1806}{3(2458)}, 14.6^{\circ}\)

b \(\tan \theta=\dfrac{32}{10}, 72.6^{\circ}\)

b \(c=10 \sqrt{10} \approx 31.6, A \approx 34.7^{\circ}, B \approx 55.3^{\circ}\)

b \(a=\sqrt{256.28} \approx 16.0, A \approx 56.5^{\circ}, B \approx 33.5^{\circ}\)

b \(\tan ^{-1}\left(\dfrac{26}{30}\right) \approx 40.9^{\circ}, \quad 91 \sqrt{1676} \approx 3612.6 \mathrm{~cm}\)

49. (a) and (b)

51. (a) and (d)

53. \(\dfrac{\sqrt{3}}{2} \approx 0.8660\)

55. \(\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3} \approx 0.5774\)

63. \(a=3 \sqrt{3}, b=3, B=30^{\circ}\)

65. \(a=b=4 \sqrt{2}, B=45^{\circ}\)

67. \(e=4, f=4 \sqrt{3}, F=120^{\circ}\)

69. \(d=2 \sqrt{3}, e=2 \sqrt{2}, f=\sqrt{2}+\sqrt{6}, F=75^{\circ}\)

71. \(\a=20, b=20, c=20 \sqrt{2})

a \(32 \sqrt{3} \mathrm{~cm}\)

b \(128 \sqrt{3} \mathrm{sq} \mathrm{cm}\)

a \(10 \mathrm{sq} \mathrm{cm}\)

b \(10 \sqrt{2} \mathrm{sq} \mathrm{cm}\)

c \(10 \sqrt{3} \mathrm{sq} \mathrm{cm}\)

2.4 Chapter 2 Summary and Review

Chapter 2 review problems.

1. If \(C>93^{\circ}\), then \(A+B+C>180^{\circ}\)

3. If \(A<B<58^{\circ}\), then \(A+B+C<180^{\circ}\)

5. If \(C>50^{\circ}\), then \(A+B+C>180^{\circ}\)

9. \(a = 97\)

11. \(c = 52\)

15. \(\theta=35.26^{\circ}\)

17. No. \(a=6, c=10\) or \(a=9, c=15\)

a \(w=86.05\)

b \(\sin \theta=0.7786, \quad \cos \theta=0.6275, \quad \tan \theta=1.2407\)

a \(y=16.52\)

b \(\sin \theta=0.6957, \quad \cos \theta=0.7184, \quad \tan \theta=0.9684\)

23. \(a = 7.89\)

25. \(x = 3.57\)

27. \(b = 156.95\)

29. \(A=30^{\circ}, a=\dfrac{23 \sqrt{3}}{3}, c=\dfrac{46 \sqrt{3}}{3}\)

31. \(F=105^{\circ}, d=10 \sqrt{2}, e=20, f=10+10 \sqrt{3}\)

35. 43.30 cm

37. 15.92 m

39. \(114.02 \mathrm{ft}, 37.87^{\circ}\)

a \(60.26^{\circ}\)

b \(60.26^{\circ}\)

c \(m=\dfrac{7}{4}=\tan \theta\)

b \(b-a,(b-a)^2\)

c \(\dfrac{1}{2} a b\)

d \(4\left(\dfrac{1}{2} a b\right)+(a-b)^2=2 a b+b^2-2 a b+a^2=a^2+b^2\)