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Trigonometry Worksheets
Free worksheets with answer keys.
Enjoy these free sheets. Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. Plus each one comes with an answer key.
(This sheet is a summative worksheet that focuses on deciding when to use the law of sines or cosines as well as on using both formulas to solve for a single triangle's side or angle)
 Law of Sines
 Ambiguous Case of the Law of Sines
 Law Of Cosines
 Sine, Cosine, Tangent, to Find Side Length
 Sine, Cosine, Tangent Chart
 Inverse Trig Functions
 Real World Applications of SOHCATOA
 Mixed Review
 Vector Worksheet
 Unit Circle Worksheet
 Graphing Sine and Cosine Worksheet
Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!
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Trigonometry Examples
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5.2 Solve Applications: Sine, Cosine and Tangent Ratios.
Learning Objectives
By the end of this section, you will be able to:
 Find missing side of a right triangle using sine, cosine, or tangent ratios
 Find missing angle of a right triangle using sine, cosine, or tangent ratios
 Solve applications using right angle trigonometry
Sine, Cosine, and Tangent Ratios
We know that any right triangle has three sides and a right angle. The side opposite to the right angle is called the hypotenuse. The other two angles in a right triangle are acute angles (with a measure less than 90 degrees). One of those angles we call reference angle and we use θ (theta) to represent it.
The hypotenuse is always the longest side of a right triangle. The other two sides are called opposite side and adjacent side. The names of those sides depends on which of the two acute angles is being used as a reference angle.
In the right triangle each side is labeled with a lowercase letter to match the uppercase letter of the opposite vertex.
Label the sides of the triangle and find the hypotenuse, opposite, and adjacent.
We labeled the sides with a lowercase letter to match the uppercase letter of the opposite vertex.
c is hypotenuse
a is opposite
b is adjacent
Label the sides of the triangle and find the hypotenuse, opposite and adjacent.
y is hypotenuse
z is opposite
x is adjacent
r is hypotenuse
t is opposite
s is adjacent
Trigonometric Ratios
Trigonometric ratios are the ratios of the sides in the right triangle. For any right triangle we can define three basic trigonometric ratios: sine, cosine, and tangent.
Let us refer to Figure 1 and define the three basic trigonometric ratios as:
Three Basic Trigonometric Ratios
Where θ is the measure of a reference angle measured in degrees.
Very often we use the abbreviations for sine, cosine, and tangent ratios.
Some people remember the definition of the trigonometric ratios as SOH CAH TOA.
For the given triangle find the sine, cosine and tangent ratio.
First let’s label the sides of the triangle:
For the given triangle find the sine cosine and tangent ratio.
For the given triangle find the sine, cosine and tangent ratio.
When calculating we will usually round the ratios to four decimal places and at the end our final answer to one decimal place unless stated otherwise.
For the given triangle find the sine, cosine and tangent ratios. If necessary round to four decimal places.
We have two possible reference angles: R and S.
Using the definitions, the trigonometric ratios for angle R are:
sin R=  cos R=  tan R= 
Using the definitions, the trigonometric ratios for angle S are:
sin S =  cos S =  tan S = 
For the given triangle find the sine, cosine, and tangent ratios. If necessary round to four decimal places.
sin F =  cos F =  tan F = 
sin D =  cos D =  tan D = 
For given triangle find the sine, cosine and tangent ratios. If necessary round to four decimal places.
sin A =  cos A =  tan A = 
sin C =  cos C =  tan C = 
Now, let us use a scientific calculator to find the trigonometric ratios. Can you find the sin, cos, and tan buttons on your calculator? To find the trigonometric ratios make sure your calculator is in Degree Mode.
Using a calculator find the trigonometric ratios. If necessary, round to 4 decimal places.
Make sure your calculator is in Degree Mode. Using a calculator find:
a) sin 30° = 0.5
b) cos 45° = 0.7071 Rounded to 4 decimal places.
c) tan 60° = 1.7321 Rounded to 4 decimal places.
Find the trigonometric ratios. If necessary, round to 4 decimal places.
a) sin 60° = 0.8660
b) cos 30° = 0.8660
c) tan 45° = 1
a) sin 35° = 0.5736
b) cos 67 ° = 0.3907
c) tan 83° = 8.1443
Finding Missing Sides of a Right Triangle
In this section you will be using trigonometric ratios to solve right triangle problems. We will adapt our problem solving strategy for trigonometry applications. In addition, since those problems will involve the right triangle, it is helpful to draw it (if the drawing is not given) and label it with the given information.We will include this in the first step of the problem solving strategy for trigonometry applications.
HOW TO: Solve Trigonometry Applications
 Read the problem and make sure all the words and ideas are understood. Draw the right triangle and label the given parts.
 Identify what we are looking for.
 Label what we are looking for by choosing a variable to represent it.
 Find the required trigonometric ratio.
 Solve the ratio using good algebra techniques.
 Check the answer by substituting it back into the ratio in step 4 and by making sure it makes sense in the context of the problem.
 Answer the question with a complete sentence.
In the next few examples, having given the measure of one acute angle and the length of one side of the right triangle, we will solve the right triangle for the missing sides.
Find the missing sides. Round your final answer to two decimal places
1. the problem and make sure all the words and ideas are understood. Draw the right triangle and label the given parts.  A drawing is given. Angle Y is our reference angle, y is opposite side, z is adjacent side, and x=14 is the hypotenuse.  
2. what we are looking for.  a) the opposite side  b) adjacent side 
3. what we are looking for by choosing a variable to represent it.  y=?  z=? 
4. the required trigonometric ratio.  sin 35° =  cos 35° = 
5. the ratio using good algebra techniques.  14 sin 35° = y 8.03 = y  14 cos 35° = z 11.47 = z 
6. the answer in the problem and by making sure it makes sense.  0.57 0.57 = 0.57  0.82 0.82 = 0.82 
7. the question with a complete sentence.  The opposite side is 8.03  The adjacent side is 11.47 
Find the missing sides. Round your final answer to one decimal place.
Find the hypotenuse. Round your final answer to one decimal place.
1. the problem and make sure all the words and ideas are understood. Draw the right triangle and label the given parts.  A drawing is given. Angle S is our reference angle, s is opposite side, r = 4 is the adjacent side, and p is the hypotenuse 
2. what we are looking for.  the hypotenuse 
3. what we are looking for by choosing a variable to represent it.  p=? 
4. the required trigonometric ratio.  cos 32° = 
5. the ratio using good algebra techniques.  0.8480 = p = 4.7170 Rounding the ratios to 4 decimal places 
6. the answer in the problem and by making sure it makes sense.  0.8480 0.8480 = 0.8480 
7. the question with a complete sentence.  The hypotenuse is 4.7 Round my final answer to one decimal place. 
Finding Missing Angles of a Right Triangle
Sometimes we have a right triangle with only the sides given. How can we find the missing angles? To find the missing angles, we use the inverse of the trigonometric ratios. The inverse buttons sin 1 , cos 1 , and tan 1 are on your scientific calculator.
Find the angles. Round your final answer to one decimal place.
a) sin A = 0.5
b) cos B = 0.9735
c) tan C = 2.89358
Use your calculator and press the 2nd FUNCTION key and then press the SIN, COS, or TAN key
a) A = sin 1 0.5
b) B = cos 1 0.9735
c) C = tan 1 2.89358
a) sin X = 1
b) cos Y = 0.375
c) tan Z = 1.676767
a) sin C = 0
b) cos D = 0.95
c) tan F = 6.3333
In the example below we have a right triangle with two sides given. Our acute angles are missing. Let us see what the steps are to find the missing angles.
1. the problem and make sure all the words and ideas are understood. Draw the right triangle and label the given parts.  A drawing is given. Angle T is our reference angle, t = 7 is the opposite side, s is adjacent side, and r =11 is the hypotenuse 
2. what we are looking for.  angle T 
3. what we are looking for by choosing a variable to represent it.  
4. the required trigonometric ratio.  sin T = 
5. the ratio using good algebra techniques.  sin T = 0.6364 T = sin 0.6364

6. the answer in the problem and by making sure it makes sense.  sin 39.5239° 0.6364 = 0.6364 
7. the question with a complete sentence.  The missing angle T is 39.5°. 
Find the missing angle X. Round your final answer to one decimal place.
Find the missing angle Z. Round your final answer to one decimal place.
Find the missing angle A. Round your final answer to one decimal place.
1. the problem and make sure all the words and ideas are understood. Draw the right triangle and label the given parts.  A drawing is given. Angle A is our reference angle, a = 9 is the opposite side, c = 5 is the adjacent side, and b is the hypotenuse 
2. what we are looking for.  angle A 
3. what we are looking for by choosing a variable to represent it.  
4. the required trigonometric ratio.  tan A = 
5. the ratio using good algebra techniques.  tan A = 1.8 A = tan 1.8

6. the answer in the problem and by making sure it makes sense.  tan 60.9° 1.8 = 1.8 
7. the question with a complete sentence.  The missing angle A is 60.9°. 
Find the missing angle C. Round your final answer to one decimal place.
Find the missing angle E. Round your final answer to one decimal place.
Solving a Right Triangle
From the section before we know that any triangle has three sides and three interior angles. In a right triangle, when all six parts of the triangle are known, we say that the right triangle is solved.
Solve the right triangle. Round your final answer to one decimal place.
Since the sum of angles in any triangle is 180°, the measure of angle B can be easy calculated.
1. the problem and make sure all the words and ideas are understood. Draw the right triangle and label the given parts.  A drawing is given. Angle A is our reference angle, a = 8 is the opposite side, b is the adjacent side, and c is the hypotenuse.  
2. what we are looking for.  a) adjacent side  b) hypotenuse 
3. what we are looking for by choosing a variable to represent it.  b = ?  c = ? 
4. the required trigonometric ratio.  tan 42° =  sin 42° = 
5. the ratio using good algebra techniques.  0.9004 = 0.9004 b = 8 b = 8.8849  0.6691 = 0.6691 c = 8 c = 11.9563 
6. the answer in the problem and by making sure it makes sense.  tan 42 ° 0.9 = 0.9  sin 42° 0.6691 = 0.6691 
7. the question with a complete sentence.  The adjacent side is 8.9. Rounded to one decimal place.  The hypotenuse is 12 
We solved the right triangle
a = 8  b = 8.9  c = 12 
TRY IT 10.1
a = 6  b = 15.6  c = 16.7 
TRY IT 10.2
a = 2.9  b = 10  c = 10.4 
Solve the right triangle. Round to two decimal places.
1. the problem and make sure all the words and ideas are understood. Draw the right triangle and label the given parts.  A drawing is given. Let angle D be our reference angle, d = 4 is the opposite side, f is the adjacent side, and e = 9 is the hypotenuse  
2. what we are looking for.  a) angle D  b) adjacent 
3. what we are looking for by choosing a variable to represent it.  f = ?  
4. the required trigonometric ratio.  sin D =  4 + f = 9 
5. the ratio using good algebra techniques.  sin D = 0.4444 D = sin 0.4444
 16 + f = 81 f = 81 – 16 f = 65 f = square root of 65 f = 8.06 
6. the answer in the problem and by making sure it makes sense.  sin 26.3850° 0.4444 =0.4444  4 + 8.06 81 = 81 
7. the question with a complete sentence.  The missing angle D is 26.39°.  The adjacent side is 8.06 Rounded to two decimal places 
The missing angle F = 180° – 90° – 26.39° = 63.61°
d = 4  e = 9  f = 8.06 
TRY IT 11.1
d = 9  e = 18.4  f = 16 
TRY IT 11.2
Solve the right triangle. Round to one decimal place.
d = 7.1  e = 10  f = 7 
Solve Applications Using Trigonometric Ratios
In the previous examples we were able to find missing sides and missing angles of a right triangle. Now, let’s use the trigonometric ratios to solve reallife problems.
Many applications of trigonometric ratios involve understanding of an angle of elevation or angle of depression.
The angle of elevation is an angle between the horizontal line (ground) and the observer’s line of sight.
The angle of depression is the angle between horizontal line (that is parallel to the ground) and the observer’s line of sight.
James is standing 31 metres away from the base of the Harbour Centre in Vancouver. He looks up to the top of the building at a 78° angle. How tall is the Harbour Centre?
1. the problem and make sure all the words and ideas are understood. Draw the right triangle and label the given parts.  Angle X is our reference angle, x is opposite side, y = 31 m is the adjacent side, and z is the hypotenuse. 
2. what we are looking for.  The opposite side 
3. what we are looking for by choosing a variable to represent it.  x=? 
4. the required trigonometric ratio.  tan 78° = 
5. the ratio using good algebra techniques.  4.7046 = x = 145.8426 
6. the answer in the problem and by making sure it makes sense.  4.7046 4.7046 = 4.7046 
7. the question with a complete sentence.  The Harbour Centre is 145.8426 metres or rounded to 146 metres. 
TRY IT 12.1
Nicole is standing 75 feet away from the base of the Living ShangriLa, the tallest building in British Columbia. She looks up to the top of the building at a 83.5° angle. How tall is the Living ShangriLa?
658.3 feet.
TRY IT 12.2
Kelly is standing 23 metres away from the base of the tallest apartment building in Prince George and looks at the top of the building at a 62° angle. How tall is the building?
43.3 metres
Thomas is standing at the top of the building that is 45 metres high and looks at his friend that is standing on the ground, 22 metres from the base of the building. What is the angle of depression?
1. the problem and make sure all the words and ideas are understood. Draw the right triangle and label the given parts. 
Angle Y is our reference angle, y = 45 m is the opposite side, z = 22 m is the adjacent side, and x is the hypotenuse 
2. what we are looking for.  angle Y 
3. what we are looking for by choosing a variable to represent it.  
4. the required trigonometric ratio.  tan Y = 
5. the ratio using good algebra techniques.  tan Y = 2.0455 Y = tan ¹2.0455

6. the answer in the problem and by making sure it makes sense.  tan 63.9470° 2.0455 = 2.0455 
7. the question with a complete sentence.  The angle of depression is 63.9470° or 64° rounded to one decimal place. 
TRY IT 13.1
Hemanth is standing on the top of a cliff 250 feet above the ground and looks at his friend that is standing on the ground, 40 feet from the base of the cliff. What is the angle of depression?
TRY IT 13.2
Klaudia is standing on the ground, 25 metres from the base of the cliff and looks up at her friend on the top of a cliff 100 metres above the ground. What is the angle of elevation?
Key Concepts
 Check the answer by substituting it back into the ratio solved in step 5 and by making sure it makes sense in the context of the problem.
Practice Makes Perfect
Label the sides of the triangle.
1
 2.

3. If the reference angle in Question 1 is B, Find the adjacent ?
 4. If the reference angle in Question 2 is Z, find the opposite ? 
5.  6. 
Use your calculator to find the given ratios. Round to four decimal places if necessary:
7.  8. 
9.  10. 
For the given triangles, find the sine, cosine and tangent of the θ.
11.  12. 
13.  14. 
For the given triangles, find the missing side. Round it to one decimal place.
15. Find the hypotenuse.  16. Find b. 
17. Find the opposite.  18. Find the adjacent. 
For the given triangles, find the missing sides. Round it to one decimal place.
19.  20. 
Solve the triangles. Round to one decimal place.
21.  22. 
23.  24. 
25. A surveyor stands 75 metres from the bottom of a tree and looks up at the top of the tree at a 48° angle. How tall is the tree?  26. A tree makes a shadow that is 6 metres long when the angle of elevation to the sun is 52°. How tall is the tree? 
27. A ladder that is 15 feet is leaning against a house and makes a 45° angle with the ground. How far is the base of the ladder from the house?  28. Matt is flying a kite and has let out 100 feet of string. The angle of elevation with the ground is 38°. How high is his kite above the ground? 
29. Marta is flying a kite and has let out 28 metres of string. If the kite is 10 metres above the ground, what is the angle of elevation?  30. An airplane takes off from the ground at the angle of 25°. If the airplane traveled 200 kilometres, how high above the ground is it? 
1.
 3. c  5.
g is opposite , f is adjacent, and e is hypotenuse 
7. 0.7314  9. 0.2126  11.
sin θ = 
13. sin θ =  15. b = 19.8  17. c = 12 
19. y = 19.3, z = 8.2  21.
b = 38.5 c = 21.3 d = 44  23.
t = 15 r = 25 s = 20 
25. 83.3 m  27. 10.6 ft  29. 20.9° 
Intermediate Algebra I Copyright © 2021 by Pooja Gupta is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.
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Trigonometry Questions and Answers
In this blog on trigonometry, we will solve some trigonometry questions and answers to brush up on our concept using identities. Trigonometry is a combination of two Greek words ‘ Trigonon ’ meaning a triangle and ‘ metron ’ meaning measure. In this blog, we will solve some of the problem statements to understand this concept better. Test your understanding of the practice problem and solve the stepbystep solution.
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Problems on Trigonometric Identities  Word Problems Involving Trigonometric Identities
Check problems on trigonometric identities along with the solutions. Find the step by step procedure to know the trigonometric identities problems. Refer to all the solutions present in the below sections to prepare for the exam. Scroll down the page to get the Trigonometric Identities Word Problems and study material. Know the various formulae involved in solving trigonometric identities below. Assess your knowledge level taking the help of the Practice Problems on Trigonometric Identities available.
Also, Read:
 Worksheet on Trigonometric Identities
 Trigonometrical Ratios Table
The angle of inclination from a point on the ground 30 feet away to the top of Lakeland’s Armington Clocktower is 60°. Find the height of the clock tower which is nearest to the foot?
As given in the question,
The angle of inclination (θ) = 60°
The height from the ground(a) = 30 feet
To find the height of the Clocktower to the nearest foot, we use the formula
tan θ = h/a
tan 60° = h / 30
h = 30 tan 60°
h = 9.60 ≅ 10 feet
Therefore, the height of the clocktower to the nearest foot is 10 feet.
Hence, the final solution is 10 feet.
Mary wants to determine the California redwood tree height, there are two sightings available from the ground which one is 200 feet directly behind the other. If the angles of inclination (Θ) are 45° and 30° respectively, how tall is the tree to the nearest foot?
Length at which trees are slighting = 200
The angle of inclinations = 45° and 30°
To find the inclination on the first tree, we apply the Pythagorean theorem,
tan 45° = h/x
h = x tan 45° is the (1) equation
tan 30° = h/(200+x)
h = (200 + x) tan 30 is the (2) equation
From both the equations,
x tan 45° = (200 + x) tan 30°
x tan 45° = 200 tan 30° + x tan 30°
x tan 45° – x tan 30° = 200 tan 30
Divide the equation with tan 45° – tan 30°
x (tan 45° – tan 30°) / tan 45° – tan 30° = 200 tan 30° / tan 45° – tan 30°
x = 273.21 feet
h = 273.21 tan 45
h = 273. 21 feet
h ≅ 273 feet
Therefore, the height of the tree to the nearest foot = 273 feet
Thus, the final solution = 273 feet
A tree that is standing vertically on the level ground casts the 120 foot long shadow. The angle of elevation from the end of the shadow of the top of the tree is 21.4°. Find the height of the tree to the nearest foot?
Length of the footlong shadow = 120
The inclination of the tree = 21.4°
To find the height of the tree to the nearest foot, we apply the Pythagorean theorem
tan θ = 0/a
tan 214° = h/120
h = 120 tan 214°
h ≅ 47 feet
The height of the tree to the nearest foot = 47 feet
Thus, the final solution is 47 feet
The broadcast tower which is for the radio station WSAZ (“Carl” and “Jeff”‘s home of algebra) has 2 enormous flashing red lights on it. Of the 2 enormous flashing lights, one is at the very top and the other one is few feet below the top. From that point to the base of the tower it is 5000 feet away from level ground, the top light angle of elevation is 7.970° and the light angle of elevation of the second light is 7.125°. Find the distance between the nearest foot and the lights.
Height of the tower = 5000 feet
The angle of elevation of top light = 7.970
The angle of elevation of second light = 7.125°
To find the distance between the nearest foot and the lights, we have to use the Pythagorean theorem
tan θ = h/5000
h = 5000 tan 7.97°
tan β = hx/5000
hx = 5000 tan (7.125)°
x = h – 5000 tan (7.125)°
x = 5000 tan 7.97° – 5000 tan 7.125°
x = 5000 (tan 7.97° – tan 7.125°)
x = 75.04 feet
x ≅ 75 feet
Therefore, 75 feet is the distance between the nearest foot and the lights
Thus, the final solution is 75 feet.
Find the solution of tan (θ) = sin (θ) sec (θ)?
sinθ/ cosθ = sinθ secθ
sinθ. (1/cosθ) = sinθ secθ
sinθ secθ = sinθ secθ
tanθ = sinθ secθ
= sinθ(1/cosθ)
= sinθ/cosθ
∴ Hence it is proved
Prove that (sec(θ)tan(θ))(sec(θ)+tan(θ))=1
sec²(θ)tan²(θ)=1
1/cos²(θ)sin²(θ)/cos(θ)=1
(1sin²(θ))/cos²(θ)=1
cos²(θ)/cos²(θ)=1
(1sin²(θ))/cos²(θ)
1/cos²(θ)sin²(θ)/cos²(θ)=sec²(θ)tan²(θ)
(sec(θ)tan(θ))(sec(θ)+tan(θ))
Therefore, (sec(θ)tan(θ))(sec(θ)+tan(θ)) = 1
∴Hence, it is proved
Prove that sec(θ)/(1tan(θ))=1/(cos(θ)sin(θ))
1/((cos(θ)sin(θ)).1/cos(θ).1/cos(θ)
(1/cos(θ))/((cos(θ)sin(θ))(cos(θ))=sec(θ)/(1tan(θ))
Therefore, sec(θ)/(1tan(θ))=1/(cos(θ)sin(θ))
Problem 8 :
An aeroplane over the Pacific sights an atoll at a 20° angle of depression. If the plane is 435 ma above water, how many kilometres is it from a point 435m directly above the centre of a troll?
The angle of depression = 20°
Height of the plane above water = 435ma
Height above the centre of a troll = 435m
To find the kilometers, we use the pythegorean theorem
tan 20° = 435/x
x = 435/tan20°
x = 1.195 km
Therefore, 1.195 kilometres is it from a point 435m directly above the centre of a troll
Thus, the final solution is 1.195 kilometres
The force F (in pounds)on the back of a person when he or she bends over an acute angle θ (in degrees) is given by F = 0.2W sin(θ + 70)/sin12° where w is the weight in pounds of the person
a) Simplify the formula or F.
b) Find the force on the back of a person where an angle of 30° weight is 50 pounds if he bends an angle of 30°
c) How many pounds should a person weigh for his book to endure a force of 400 lbs if he bends 40°?
a. F = 0.2W sin (θ + 90)/sin 12°
= 0.2W [sinθ cos 90 + cosθ sin90]/sin 12°
= 0.2W [θ(sinθ) + (cosθ) (1)]/sin 12°
= 0.2W [0 + cosθ]/sin 12°
F = 0.2W(cosθ)/sin 12°
The value of F is 0.2W(cosθ)/sin 12°
b. W = 50, θ = 30°, F=?
F = 0.2W cosθ/sin 12°
F = 0.2 (50) (cos30°)/sin 12°
The force on the back of a person wherein the angle of 30e weight is 50 pounds if he bends an angle of 30° is 41.45
c. F = 400, θ = 40°, W = ?
400 = 0.2(w)(cos 40°)/sin 12°
400(sin 12°)/0.2 cos 40 = 0.2 (w) (cos 40°)/0.2 cos40°
Therefore, the weigh for his book to endure a force of 400 lbs if he bends 40° is 542.82 pounds
Problem 10:
An observer standing on the top of vertical cliff pots a house in the adjacent valley at an angle of depression of 12°. The cliff is 60m tall. How far is the house from the base of the cliff?
The angle of depression = 12°
Height of the cliff = 60m
To find, the distance of the house from the base of the cliff, we apply the Pythagorean theorem
tan 12° = 60/x
x = 60/tan 12°
282m is the distance of the house from the base of the cliff
Hence, the final solution is 282m
Problem 11:
Building A and B are across the street from each other which is 35m apart. From the point on the roof of building A, the angle of elevation of the top of building B is 24°, the angle of depression of the base of building B is 34° How tall is each building?
The angle of elevation of the top of the building = 24°
The angle of depression of the base of the building = 34°
The distance of both buildings = 35m
To find the height of each building, we apply the Pythagoras theorem,
tan 24° = c/35
tan 56° = 35/a
A is 23.6m tall
B is 39.2m tall
Therefore, the height of building A is 23.6m tall
The height of building B is 39.2m tall
Thus the final solution is 23.6m, 39.2m
Problem 12:
In Johannesburg in June, the daily low temperature is usually around 3°C, and the daily temperature is around 18°. The temperature is typically halfway between the daily high and daily low at 10 am and 10 pm. and the highest temperatures are in the afternoon. Find out the trigonometric function which models the temperature T in Johannesburg t hours after midnight?
As per the question,
To determine the trigonometric model, the temperature ‘T’ which is Celsius degree and the temperature in axis and then right over here is time in hours. To think about the range of temperatures, the daily temperature is around 3degree celsius and the highest is 18°. The midpoint between 18 and 3 is 10.5 (21 divided by 2)
Let F(t) is the temperature t hours after 10 am
F(t) = 7.5sin(2Π/24 t) + 10.5
T (t) = 7.5sin (Π/12(t10)) + 10.5
T(10) = F(0) where T(10) is temperature at 10 pm
F(0) is the temperature at 10 am
Therefore, T(10) = F(0) is the trigonometric function that models the temperature T in Johannesburg t hours after midnight
Problem 13:
A ladder is 6 meters long and reaches the wall at a point of 5m from the ground. What is the angle which the ladder will make with the wall?
Let θ be the inclination of the ladder which it makes with the wall
Length of the ladder = 6m
The distance at which the ladder touches the wall = 5m
The angle at which the ladder will make with the wall
cos θ = 5/6
θ = cos¯¹ (5/6)
Problem 14:
The acceleration of the piston is given by a = 5.0(sinωt + cos2ωt). At what positive values of ωt less than 2Π does a = 0?
Let ωt be the crank angle(product of time and angular velocity) in piston acceleration, a be the accelaration of a piston.
a = 5.0(sinωt + cos2ωt)
0 = 5.0(sinωt + cos2ωt)
0 = sinωt + cos2ωt
0 = sinωt + 1 – 2sin²ωt
2sin²ωt – sinωt – 1 = 0
(2sinωt + 1) (sinωt – 1) = 0
2sinωt + 1 = 0
2sinωt = 1
sinωt = 1/2
ωt = sin¯¹(1/2)
ωt = (7π/6, 11π/6)
sinωt – 1 = 0
ωt = sin¯¹(1)
{π/2, 7π/6, 11π/6}
Problem 15:
A lighthouse at sea level is 34 mi from a boat. It is known that the top of the lighthouse is 42.5mi from the boat. Find the angle of depression from the top of the lighthouse.
Let θ be the angle of inclination from the top of the lighthouse to the boat
Let x be the horizontal distance from the base of the lighthouse to the boat, and
r be the distance from the top of the lighthouse to the boat.
The length of the lighthouse at sea level = 34 mi
The distance of the top lighthouse from the boat = 42.5 mi
To find the angle of depression from the top of the lighthouse, we apply the Pythagorean theorem
cosθ = 34/42.5
θ = cos¯¹(34/42.5)
Number Line
 \mathrm{Lauren's\:age\:is\:half\:of\:Joe's\:age.\:Emma\:is\:four\:years\:older\:than\:Joe.\:The\:sum\:of\:Lauren,\:Emma,\:and\:Joe's\:age\:is\:54.\:How\:old\:is\:Joe?}
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 \mathrm{The\:sum\:of\:two\:numbers\:is\:249\:.\:Twice\:the\:larger\:number\:plus\:three\:times\:the\:smaller\:number\:is\:591\:.\:Find\:the\:numbers.}
 \mathrm{If\:2\:tacos\:and\:3\:drinks\:cost\:12\:and\:3\:tacos\:and\:2\:drinks\:cost\:13\:how\:much\:does\:a\:taco\:cost?}
 \mathrm{You\:deposit\:3000\:in\:an\:account\:earning\:2\%\:interest\:compounded\:monthly.\:How\:much\:will\:you\:have\:in\:the\:account\:in\:15\:years?}
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