# Trigonometry Questions and Answers updated daily – Aptitude

Trigonometry Questions: Solved 151 Trigonometry Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Trigonometry Questions

1. If 2ycosÎ¸ = xsinÎ¸ and 2xsecÎ¸ - ycosecÎ¸ = 3, then value of x2 + y2 is

Correct Ans:4
Explanation:
Given:
2ycosÎ¸ = xsinÎ¸ ; 2xsecÎ¸ - ycosecÎ¸ = 3

2xsecÎ¸ - ycosecÎ¸ = 3
2x/cosÎ¸ - y/sinÎ¸ = 3
2xsinÎ¸ - ycosÎ¸ = 3sinÎ¸ cosÎ¸ ..(i)

2ycosÎ¸ = xsinÎ¸
ycosÎ¸ = xsinÎ¸/2

Substitute ycosÎ¸ in (i),
2xsinÎ¸ - (xsinÎ¸/2) = 3sinÎ¸ cosÎ¸
(4xsinÎ¸ - xsinÎ¸)/2 = 3sinÎ¸ cosÎ¸
4xsinÎ¸ - xsinÎ¸ = 6sinÎ¸ cosÎ¸
3xsinÎ¸ = 6sinÎ¸ cosÎ¸
x = 2cosÎ¸

ycosÎ¸ = 2cosÎ¸sinÎ¸/2
y = sinÎ¸
x2 + y2 = 4cos2Î¸ + sin2Î¸
= 4(1) (wkt, sin2Î¸ + cos2Î¸ = 1 )
= 4.
Workspace

2. If θ be acute angle and cosθ = 15/17, then the value of cot (90° - θ) is

Correct Ans:8/15
Explanation:
Given:
Cosθ = 15/17
WKT, sinθ = √(1 - cos2θ)
sinθ = √(1 - (15/17)2)
sinθ = √( 1 - 225/289)
sinθ = √( 64/289)
sinθ = 8/17

Cot (90°- θ) = tanθ
So, tanθ = sinθ/cosθ = (8/17)/(15/17)
= 8/15.
Workspace

3. If cotα = 3, then the value of (sin3α + cos3α)/cosα is ?

Correct Ans:14/15
Explanation:
Given,
cotα = 3
cosα/sinα = 3
cosα = 3sinα
(sin3α + cos3α)/cosα
= (sin3α + 27sin3α)/3sinα
= 28sin3α/3sinα
= 28sin2α/3
WKT, 1/sinα = cosecα
= 28/3cosec2α
As, cosec2α - cot2α = 1
= (28/3)*(1/(1 + cot2α))
= (28/3)*(1/(1 + 32))
= (28/3)*(1/(1 + 9)
= (28/3)*(1/10)
= 14/15.
Workspace

4. If 7sinÎ± = 24cosÎ±; 0 < Î± < Ï€/2, then the value of 14tanÎ± - 75cosÎ± - 7secÎ± is equal to

Correct Ans:2
Explanation:
Given: 7sinÎ± = 24cosÎ±
Here, 24 --- opposite side
So, Hypotenuse = √[opposite side2 + adjacent side2]
Hypotenuse = √[242 + 72]
= √[576 + 49]
= √625
= 25

cosÎ± = adjacent side/hypotenuse = 7/25
secÎ± = 1/cosÎ± = 25/7

14tanÎ± - 75cosÎ± - 7secÎ± = 14(24/7) - 75(7/25) - 7(25/7)
= 48 - 21 - 25
= 2.
Workspace

5. Provided sin (A "“ B) = sinA cosB "“ cosA sinB, then sin 15Â° will be

Correct Ans:(√3 - 1)/2√2
Explanation:
sin 15Â° = sin (45Â° â€“ 30Â°)
= sin 45Â° . cos 30Â° â€“ cos 45Â°. sin 30Â°
= (1/√2) x (√3/2) - (1/√2) x (1/2)
= (√3 /2√2) - (1/2√2)
= (√3 - 1)/2√2.
Workspace

6. The angles of elevation of the top of a building from the top and bottom of a tree are x and y respectively. If the height of the tree is h metre, then, in metre, the height of the building is.

Correct Ans:h cotx/(cotx - coty)
Explanation:

In âˆ†ABE,
tan x = x/BE
BE = x/tan x ....(i)

tan y = (x + h)/DC
DC = (x + h)/tan y ...(ii)
From (i) & (ii),
Since, DC = BE
tan y = (x + h)/(x/tan x)
tan y(x/tan x) = x + h
x tan y cot x = x + h
x(tan y cot x - 1) = h
x = h/(tan y cot x - 1)

Height of the building = x + h
= {h/(tan y cot x - 1)} + h
= {h/[(cot x/cot y) - 1]} + h
= {h cot y/(cot x - cot y)} + h
= [h cot y + h cot x - h cot y]/(cot x - cot y)
= h cot x/(cot x - cot y).
Workspace

7. If sin 21Â° = x/y, then sec 21Â° - sin 69Â° is equal to

Correct Ans:x2/y√(y2 - x2)
Explanation:
Given:
sin 21Â° = x/y
sin(90Â° - 69Â°) = x/y
cos 69Â° = x/y
sin 69Â° = √(1 - cos2 69Â°)
= √(1 - (x/y)2)
= √(y2 - x2)/y

Similarly,
cos 21Â° = √(1 - sin2 21Â°)
= √(1 - (x/y)2)
= √(y2 - x2)/y

sec 21Â° - sin 69Â° = (1/cos 21Â°) - sin 69Â°
= y/√(y2 - x2) - √(y2 - x2)/y
= (y2 - y2 + x2)/y√(y2 - x2)
= x2/y√(y2 - x2).
Workspace

8. If xsin3θ + ycos3θ = sinθ cosθ and x sinθ - y cosθ = 0, then the value of x2 + y2 equals

Correct Ans:1
Explanation:
Given:
x sinθ - y cosθ = 0
x sinθ = y cosθ ... (i)

xsin3θ + ycos3θ = sinθ cosθ
y cosθ.sin2θ + ycos3θ = sinθ cosθ
y cosθ(sin2θ + cos2θ) = sinθ cosθ
y cosθ(1) = sinθ cosθ (wkt, sin2θ + cos2θ =1)
y = sinθ

From (i),
x sinθ = y cosθ
x sinθ = sinθ cosθ
x = cosθ
Hence, x2+ y2 = cosθ2 + sinθ 2
= 1
Workspace

9. A boy standing in the middle of a field, observes a flying bird in the north at an angle of elevation of 30° and after 2 minutes, he observes the same bird in the south at an angle of elevation of 60°. If the bird flies all along in a straight line at a height of m, then its speed in km/h is

Correct Ans:6
Explanation:

When boy observes a bird at point B in north side,
In ΔABD,
1/√3 = BD/50√3
BD = 50√3/√3
BD = 50 m

After 2 minutes, when boy observes a bird at point C in south side,
In ΔACD,
√3 = CD/50√3
CD = 50√3(√3)
CD = 150 m
Total distance travelled by bird, BC = BD + CD
= 50 + 150 = 200 m
Speed of bird = Distance/Time = 200/2 = 100 m/minute
By converting m/min into km/hr,
= [100/1000]/[1/60] = 6 km/hr.
Workspace

10. If sin(x + y)/sin(x - y) = (a + b)/(a - b), then the value of tanx/tany is

Correct Ans:a/b
Explanation:
Given:
sin(x + y)/sin(x - y) = (a + b)/(a - b)
Using component and divident rule,
[sin(x + y) + sin(x - y)]/[sin(x - y) - sin(x - y)] = [a + b + a - b]/[a + b - a + b]

WKT,
Sin A + Sin B = 2[Sin(A + B)/2 * Cos(A - B)/2]
Sin A - Sin B = 2[Cos(A + B)/2 * Sin(A - B)/2]

2[sin(x + y + x - y)/2*cos(x - y - x + y)/2]/2[cos(x + y + x - y)/2*sin(x - y - x + y)/2] = 2a/2b
2sinx .cosy/2cosx .siny = a/b
tanx.coty = a/b
WKT, coty = 1/tany
tanx/tany = a/b.
Workspace

11. If sin3Î¸ sec2Î¸ = 1, then what is the value of (3tan2 (5Î¸/2) "“ 1)?

Correct Ans:2
Explanation:
Given: Sin3Î¸ sec2Î¸ = 1

WKT, secÎ¸ = 1/cosÎ¸
sin3Î¸/cos2Î¸ = 1
sin3Î¸ = cos2Î¸

WKT, cosÎ¸ = sin(90° - Î¸)
sin3Î¸ = sin(90° - 2Î¸ )
3Î¸ = 90° - 2Î¸
5Î¸ = 90°
45° = 5Î¸/2

3tan2 (5Î¸/2) â€“ 1 = 3tan245° - 1
= 3(1) - 1 ..... (wkt, tan45° = 1)
= 2.
Workspace

12. Two persons are on either side of a temple, 75 m high, observe the angle of elevation of the top of the temple to be 30Â° and 60Â° respectively. The distance between the persons is

Correct Ans:173.2 m
Explanation:

In ∆ABC,
tan 30Â° = AC/BC
1/√3 = 75/BC
BC = 75√3

In ∆ACD,
tan 60Â° = AC/CD
√3 = 75/CD
CD = 75/√3

Distance between two persons = BC + CD = 75√3 + 75/√3
= 75(√3 + 1/√3)
= 300/√3
= 173.2 m.
Workspace

13. 2 - cos2θ = 3sinθcosθ, sinθ ≠ cosθ then tanθ is

Correct Ans:1/2
Explanation:
Given:
2 - cos2θ = 3sinθcosθ
Dividing by cos2θ,
(2/cos2θ) - 1 = 3sinθcosθ/cos2θ
2sec2θ - 1 = 3tanθ
2(1 + tan2θ) - 1 = 3tanθ
2tan2θ + 2 - 1 = 3tanθ
2tan2θ -3tanθ + 1 = 0
2tan2θ - 2tanθ -tanθ + 1
2tanθ(tanθ - 1) - 1(tanθ - 1) = 0
(2tanθ - 1)(tanθ - 1) = 0
tanθ = 1/2 or 1
Here, the given option is 1/2.
Workspace

14. An person 1.8m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. Find height of the tower is?

Correct Ans:21.8m
Explanation:
Given:
Height of the person, AB = 1.8m
Angle of elevation = 30°
Distance between tower and person, AC = 20√3

From the diagram we can say,
BE = AC = 20√3
AB = CE = 1.8m

DE/BE = tan30° = 1/√3
DE/(20√3) = 1/√3
DE = 20√3/√3
DE = 20m
Height of the tower, CD = DE + CE
= 20 + 1.8
= 21.8m
Workspace

15. Ram and Shyam are 10 km apart. They both see a hot air balloon passing in the sky making an angle of 60° and 30° respectively. What is the height at which the balloon could be flying?

Correct Ans:Both a and b
Explanation:
Here, the directions are unknown so we have two different cases.
Case 1:

tan 30° = h/x
1/√3 = h/x
h = x/√3 ....(1)
tan 60° = h/(10 - x)
√3 = h/(10 - x)
h = (10 - x)√3 ....(2)
Equating (1) & (2),
(10 - x)√3 = x/√3
(10 - x)3 = x
30 - 3x = x
30 = 4x
x = 15/2
Therefore, h = x/√3 = 15/2√3
h = (5 * 3)/2√3
h = 5√3/2

Case 2:

tan 60° = h/x
√3 = h/x
h = √3 x ....(3)
tan30° = h/(10 - x)
1/√3 = h/(10 - x)
h = (10 - x)/√3 ....(4)
Equate (3) & (4),
√3 x = (10 - x)/√3
x = 10 - x
2x = 10
x = 5
Therefore, h = √3 x = 5√3
Hence, the correct option is both a and b.
Workspace

16. If cos θ + sec θ = 2, the value of cos6 θ + sec6 θ is

Correct Ans:2
Explanation:
cos θ + sec θ = 2
put θ = 0°
cos 0° + sec 0° = 2
(cos 0° = 1 and sec 0° = 1)
1 + 1 = 2
2 = 2
cos6 θ + sec6 θ
= (1)6 + (1)6
= 1 + 1 = 2
Workspace

17. If x = cosecðœƒ- sinðœƒ and y = secðœƒ - cosðœƒ, then the relation between x and y is

Correct Ans:x2y2(x2 + y2 + 3) = 1
Explanation:
Given:
x = cosecðœƒ - sinðœƒ
y = secðœƒ - cosðœƒ
Put = 45â°
x = cosecðœƒ - sinðœƒ
x = cosec(45â°) - sin(45â°)
x = √(2) - (√(2)/2)
x = 1/√(2)

y = secðœƒ - cosðœƒ
y = sec(45â°)- cos(45â°)
y = √(2) - (√(2)/2)
y = 1/√(2)
By option,
x²y²(x² + y² + 3) = (1/√(2))² * (1/√(2))² [(1/√(2))² + (1/√(2))² + 3]
=(1/2)(1/2)[(1/2) + (1/2) + 3]
= 1
This equation x²y²(x² + y² + 3) = 1 statisfies.
Workspace

18. If is an acute angle and cosðœƒ = 15/17, then the value of cot (90º - ðœƒ) is

Correct Ans:8/15
Explanation:
Given, cosðœƒ = 15/17
Secðœƒ = 1/cosðœƒ = 17/15
Cot (90º - ðœƒ) = tanðœƒ
But tanðœƒ = √sec²ðœƒ - 1 (where tan²ðœƒ = sec²ðœƒ - 1)
= √(17/15)² - 1
= √(289/225) - 1
= √(289 - 225)/225
= √64/225
= 8/15
cot (90º - ðœƒ) = 8/15
Workspace

19. If 0°<θ<90°
2sinÂ²θ + 3cosθ = 3, then the value of θ is ?

Correct Ans:60°
Explanation:
Given that,
2sinÂ²θ + 3cosθ = 3
2(1 - cosÂ²θ) + 3cosθ - 3 = 0 (where sinÂ²θ = 1 - cosÂ²θ)
2 - 2 cosÂ²θ + 3cosθ - 3 = 0
- 2 cosÂ²θ + 3cosθ - 1 = 0
2 cosÂ²θ - 3cosθ + 1 = 0
By factorization we get,
2 cosθ (cosθ - 1) - 1 (cosθ - 1) = 0
(cosθ - 1) (2cosθ - 1) = 0
If cosθ - 1
Cosθ = 1
Cosθ = Cos 0° (where cos 0° = 1)
θ = 0°
If 2cosθ - 1
2 cosθ = 1
Cosθ = 1/2
Cosθ = Cos 60° (where cos 60° = 1/2)
θ=60°
the value of θ is 60°
Workspace

20. What is the value of (2 + tan 60º) ?

Correct Ans:2 + √3
Explanation:
We know that,
tan 60º = √3
∴ 2 + tan 60º = 2 + √3
Workspace

Are you seeking for good platform for practicing Trigonometry questions in online. This is the right place. The time you spent in Fresherslive will be the most beneficial one for you.

## Online Test on Trigonometry @ Fresherslive

This page provides important questions on Trigonometry along with correct answers and clear explanation, which will be very useful for various Interviews, Competitive examinations and Entrance tests. Here, Most of the Trigonometry questions are framed with Latest concepts, so that you may get updated through these Trigonometry Online tests. Trigonometry Online Test questions are granted from basic level to complex level.

## Why To Practice Trigonometry Test questions Online @ Fresherslive?

Trigonometry questions are delivered with accurate answer. For solving each and every question, very lucid explanations are provided with diagrams wherever necessary.
Practice in advance of similar questions on Trigonometry may improve your performance in the real Exams and Interview.
Time Management for answering the Trigonometry questions quickly is foremost important for success in Competitive Exams and Placement Interviews.
Through Fresherslive Trigonometry questions and answers, you can acquire all the essential idea to solve any difficult questions on Trigonometry in short time and also in short cut method.
Winners are those who can use the simplest method for solving a question. So that they have enough time for solving all the questions in examination, correctly without any tense. Fresherslive provides most simplest methods to answer any tough questions. Practise through Fresherslive test series to ensure success in all competitive exams, entrance exams and placement tests.

## Why Fresherslive For Trigonometry Online Test Preparation?

Most of the job seekers finding it hard to clear Trigonometry test or get stuck on any particular question, our Trigonometry test sections will help you to success in Exams as well as Interviews. To acquire clear understanding of Trigonometry, exercise these advanced Trigonometry questions with answers.
You're Welcome to use the Fresherslive Online Test at any time you want. Start your beginning, of anything you want by using our sample Trigonometry Online Test and create yourself a successful one. Fresherslive provides you a new opportunity to improve yourself. Take it and make use of it to the fullest. GOODLUCK for Your Bright Future.

## Online Test

Online Test for Aptitude
Online Test for Logical Reasoning
Online Test for Computer Knowledge
Online Test for General Knowledge
Online Test for Data Interpretation
Online Test for Verbal Ability
Online Test for C++
Online Test for Networking
Online Test for Java
Online Test for C Language
FreshersLive - No.1 Job site in India. Here you can find latest 2022 government as well as private job recruitment notifications for different posts vacancies in India. Get top company jobs for both fresher and experienced. Job Seekers can get useful interview tips, resume services & interview Question and answer. Practice online test free which is helpful for interview preparation. Register with us to get latest employment news/rojgar samachar notifications. Also get latest free govt and other sarkari naukri job alerts daily through E-mail...