Trigonometry: Solved Examples - Quantitative Aptitude (Quant) - 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

Answer : 5

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

r_min = 5

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

Hence, the answer is 5.

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
Answer: 21
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?"

Hence, the answer is 21.

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
Answer: 1:2√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 = 

Answer choice (B)

The question is "What is the ratio of heights?"

Hence, the answer is 1:2√3

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
Answer: 1:2√3

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
Answer. 3
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 , 

∴ (d) Answer
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
Answer: Both A and B
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
Answer: 10 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

Answer: 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

Answer: 4√mn

Explanation-
Adding the two equations,
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
Answer: 73.2 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

Answer: √2 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.
Answer. t
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
Answer: 15√3 + 21
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.