Courses

# Trigonometry Questions with Answer Quant Notes | EduRev

## Quantitative Aptitude (Quant)

Created by: Wizius Careers

## CAT : Trigonometry Questions with Answer Quant Notes | EduRev

The document Trigonometry Questions with Answer Quant Notes | EduRev is a part of the CAT Course Quantitative Aptitude (Quant).
All you need of CAT at this link: CAT

Question 1: 3sinx + 4cosx + r is always greater than or equal to 0. What is the smallest value ‘r’ can to take?
A. 5
B. -5
C. 4
D. 3

Explanation-

3sinx + 4cosx ≥ -r 3/5 = cosA ⇒ sinA = 4/5

5(sinx cosA + sinA cosx) ≥ -r

5(sin(x + A)) ≥ -r

5sin (x + A) ≥ -r

-1 ≤ sin (angle) ≤ 1

5sin (x + A) ≥ -5

rmin = 5

The question is "What is the smallest value ‘r’ can to take?"

Choice A is the correct answer.

Question 2: Sin2014x + Cos2014x = 1, x in the range of [-5π, 5π], how many values can x take?
A. 0
B. 10
C. 21
D. 11
Explanation-
We know that Sin2x + Cos2x = 1 for all values of x.

If Sin x or Cos x is equal to –1 or 1, then Sin2014x + Cos2014x will be equal to 1.

Sin x is equal to –1 or 1 when x = –4.5π or –3.5π or –2.5π or –1.5π or –0.5π or 0.5π or 1.5π or 2.5π or 3.5π or 4.5π.

Cosx is equal to –1 or 1 when x = –5π or –4π or –3π or –2π or –π or 0 or π or 2π or 3π or 4π or 5π.

For all other values of x, Sin2014 x will be strictly lesser than Sin2x.

For all other values of x, Cos2014 x will be strictly lesser than Cos2x.

We know that Sin2x + Cos2x is equal to 1. Hence, Sin2014x + Cos2014x will never be equal to 1 for all other values of x. Thus there are 21 values.

The question is "How many values the 'x' can take?"

Choice C is the correct answer.

Question 3: Consider a regular hexagon ABCDEF. There are towers placed at B and D. The angle of elevation from A to the tower at B is 30 degrees, and to the top of the tower at D is 45 degrees. What is the ratio of the heights of towers at B and D?
A. 1:√3
B. 1:2√3
C. 1:2
D. 3:4√3
Explanation-

Let the hexagon ABCDEF be of side ‘a’. Line AD = 2a. Let towers at B and D be B’B and D’D respectively.
From the given data we know that ∠B´AB = 30° and ∠D´AB = 45°. Keep in mind that the Towers B’B and D´D are not in the same plane as the hexagon. In Triangle B’AB,
Tan∠B´AB = B´AB B’B = In Triangle D´AD, tan ∠D´AD D’D = 2a
Ratio of heights = The question is "What is the ratio of heights?"

Choice B is the correct answer.

Question 4: Find the maximum and minimum value of 8 cos A + 15 sin A + 15
A. 11√2+15
B. 30; 8
C. 32; -2
D. 23; 8

Explanation-
Always look out for Pythagorean triplets, we know that (8,15,17) is one
∴ The expression becomes: Let there be a angle B for which sin B = 8/17, cos B = 15/17
⇒ 17( sin B cos A + cos B sin A) + 15
17(sin(A+B)) + 15
We know that sin(A+B)max = 1
sin(A+B)min = -1
∴ Max value = 17 * 1 +15 = 32
Min value = 17 * -1 + 15 = -2

The question is "to find the maximum and minimum value "

##### Hence, the answer is 32; -2

Choice B is the correct answer.

Question 5: : If cos A + cos2 A = 1 and a sin12 A + b sin10 A + c sin8 A + d sin6 A - 1 = 0. Find the value of a+b/c+d

A. 4
B. 3
C. 6
D.1
Explanation.

Given,
Cos A = 1 - Cos2A
Cos A = Sin2 A
Cos2A = Sin4A
1 – Sin2 A = Sin4 A
1 = Sin4 A + Sin2 A
13 = (Sin4A + Sin2A)3
1 = Sin12 A + Sin6A + 3Sin8 A + 3Sin10 A
Sin12 A + Sin6A + 3Sin8 A + 3Sin10 A – 1 = 0
On comparing,
a = 1, b = 3 , c = 3 , d = 1 Always look out for Pythagorean triplets, we know that (8,15,17) is one The question is "Find the value of a+b/c+d"

##### Hence, the answer is 3

Choice B is the correct answer.

Question 6: In the below figure, the sheet of width W is folded along PQ such that R overlaps S Length of PQ can be written as :- A. B. C. D. Any two of the above
Answer: Any two of the above

Explanation-
If you are quick at observing , this question can be solved just by looking at the options as , Actually solving the question.
For R and S to overlap,
∆ PQR is congurent to ∆ PSQ
∴ ∠ QPR = ⁡∝ ; ∠ SQP = ∠ PQR = 90 - ⁡∝
∠ RQT = 2⁡∝ (180°-(180°-2∝))
We can draw the figure as :- ∴In ∆ QRT
QT = QR cos2⁡∝
In ∆ PSQ,
SQ = PQ sin∝ = QR
∴ w = QT + SQ = PQ sin∝cos2∝ + PQ sin∝
= PQ sin∝ (1 + cos 2∝) (as shown earlier)

The question is "Length of PQ can be written as "

##### Hence, the answer is Any two of the above

Choice D is the correct answer.

Question 7: 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?
A. B. 5√3
C. Both A and B
D. Can’t be determined
Explanation- tan 30° = h/x  h = (10 – x) √3 30 – 3x = x ⇒ x = 30/4 = 15/2 Case 2 :  Also,
tan 30° =  x = 5
H = 5√3

The question is "What is the height at which the balloon could be flying"

##### Hence, the answer is Both A and B

Choice C is the correct answer.

Question 8: A man standing on top of a tower sees a car coming towards the tower. If it takes 20 minutes for the angle of depression to change from 30° to 60°, what is the time remaining for the car to reach the tower?
A. 20√3 minutes
B. 10 minutes
C. 10√3 minutes
D. 5 minutes

Explanation-

From the figure:-  tan 60° =  h/b a + b = 3b
2b = a
⇒ b = a/2

Now since the car takes 20 minutes to travel a distance
Time taken to travel b = a/2 distance = 20/2 = 10 minutes

The question is "What is the time remaining for the car to reach the tower?"

##### Hence, the answer is 10 minutes

Choice B is the correct answer.

Question 9: A right angled triangle has a height ‘p’, base ‘b’ and hypotenuse ‘h’. Which of the following value can h2 not take, given that p and b are positive integers?
A. 74
B. 52
C. 13
D. 23

Explanation-
We know that,
h2 = p2 + b2 Given, p and b are positive integer, so h2 will be sum of two perfect squares.
We see
a) 72 + 52 = 74
b) 62 + 42 = 52
c)  32 + 22 = 13
d) Can’t be expressed as a sum of two perfect squares

The question is "Which of the following value can h2 not take, given that p and b are positive integers? "

##### Hence, the answer is 23

Choice D is the correct answer.

Question 10: tan ∅ + sin ∅ = m, tan ∅ - sin ∅ = n, Find the value of m2- n2
A. 2√mn
B. 4√mn
C. m – n
D. 2mn

Explanation-
tan ∅ = Subtracting the same,
sin ∅ = Since, there are no available direct formula for relation between sin∅ ∅tan∅ but we know that

cosec2 ∅ – cos2 ∅ = 1  ⇒((m2-n2))2 = 4(4mn)
⇒ m2 - n2 = 4√mn

The question is "To find the value of m2= n2"

##### Hence, the answer is 4√mn

Choice B is the correct answer.

Question 11: A student is standing with a banner at the top of a 100 m high college building. From a point on the ground, the angle of elevation of the top of the student is 60° and from the same point, the angle of elevation of the top of the tower is 45°. Find the height of the student.
A. 35 m
B. 73.2 m
C. 50 m
D. 75 m
Explanation- Let BC be the height of the tower and DC be the height of the student.
In rt. ∆ ABC
AB = BC cot 45°
AB = 100 m ……………(i)
In rt. ∆ ABD
AB = BD cot 60°
AB = (BC + CD) cot 60° Equating (i) and (ii) (10 + CD)= 100√3
CD = 100√3 – 100
= 10(1.732 – 1) = 100 * 0.732 = 73.2 m

The question is "To find the height of the student"

##### Hence, the answer is 73.2 m

Choice B is the correct answer.

Question 12: If Cos x – Sin x = √2 Sin x, find the value of Cos x + Sin x:
A. √2 Cos x
B. √2 Cosec x
C. √2 Sec x
D. √2 Sin x Cos x

Explanation-
Cos x – Sin x = √2 Sin x
Cos x = Sin x + √2 Sin x
Cos x = Sin x + √2 Sin x   Sin x = (√2 - 1) Cos x
Sin x = √2 Cos x – Cos x
Sin x + Cos x = √2 Cos x

The question is "To find the value of Cos x + Sin x"

##### Hence, the answer is √2 Cos x

Choice A is the correct answer.

Question 13: can be written as:
A. 1/t
B. t
C. √t Sec x
D. Explanation.  (Since, 1 = Cos2x + Sin 2x) The question is can be written as"

##### Hence, the answer is t

Choice B is the correct answer.

Question 14: A tall tree AB and a building CD are standing opposite to each other. A portion of the tree breaks off and falls on top of the building making an angle of 30°. After a while it falls again to the ground in front of the building, 4 m away from foot of the tree, making an angle of 45°. The height of the building is 6 m. Find the total height of the tree in meters before it broke.
A. 27√3 + 39
B. 12√3 + 10
C. 15√3 + 21
D. Insufficient Data
Explanation- Let the broken portion of tree AA’ be x. Hence A’C = A’G = x
From the figure, total height of the tree = x + y + 6
Consider triangle A’BG, tan 45° =  Or BG = y + 6
Consider triangle A’C’C, tan 30° =  y√3 = y + 10 Therefore y = 5(√3 + 1)
Take sin 30° to find x
sin 30° = 1/2 = y/x
or x = 2y
Height of the tree = x + y = 6
= 2 * 5(√3 + 1) + 5(√3 + 1) + 6
= 10√3 + 10 + 5√3 + 5 + 6
= 15√3 + 21 meters

The question is "To find the total height of the tree in meters before it broke"

##### Hence, the answer is 15√3 + 21

Choice C is the correct answer.

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;