Trigonometry & Height and Distances

Question 1

What is the maximum value of 3 Sinθ + 4 cosθ?

• A. 12
• B. 5
• C. 6
• D. 1

Answer: (B)

Trigonometry & Height and Distances
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Maximum value = √(a² + b²) = 5

Question 2

What is minimum value of Sinθ + cosθ ?

• A. -2
• B. √3/2
• C. -1
• D. -√2

Answer: (D)

Trigonometry & Height and Distances
Discuss it

Minimum value = √(a² - b²)
=-√2

Question 3

If tan (x+y) tan (x-y) = 1, then find tan (2x/3)?

• A. 1/√3
• B. 1/2
• C. 1/√2
• D. 2/√3

Answer: (A)

Trigonometry & Height and Distances
Discuss it

tanA = cotB, 
tanA*tan B = 1
So, A +B = 90^o
(x+y)+(x-y) = 90^o, 2x = 90^o , x = 45^o
Tan (2x/3) = tan 30^o = 1/√3

Question 4

Find the Value of tan30^o + tan120^o?

• A. √3
• B. -1/√3
• C. 0
• D. -2/√3

Answer: (D)

Trigonometry & Height and Distances
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=tan(30) + tan(180-120) =1/√3 +(-√3)  = -2/√3

Question 5

The least value of 2sin²θ + 3cos²θ

• A. 1/3
• B. 4/3
• C. 2
• D. 3/4

Answer: (C)

Trigonometry & Height and Distances
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= 2sin²θ + 2cos²θ + cos²θ
=2(sin²θ + cos²θ) + cos²θ ; (by putting sin²θ + cos²=1)
= 2 + cos²θ ;(the minimum value of cos²θ=0) = 2 + 0 = 2

Question 6

The angle of elevation of the sun, when the length of the shadow of a tree is 1/√3 times the height of the tree, is:

• A. 30 degrees
• B. 45 degrees
• C. 60 degrees
• D. 90 degrees

Answer: (C)

Trigonometry & Height and Distances
Discuss it

Let AB be the height of the tree and AC be the length of the shadow. We need to calculate the angle ACB where BC is the hypotenuse. Tanθ = Perpendicular / Base = 1/(1/√3)

Tanθ  = Tan 60°

Therefore angle of elevation is 60 deg

Question 7

From a point A on a level ground, the angle of elevation of the top of a tower is 30 degrees. If the tower is 100 m high, the distance of point A from the foot of the tower is:

• A. 148 m
• B. 156 m
• C. 173 m
• D. 200 m

Answer: (C)

Trigonometry & Height and Distances
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Let QR be the tower. Then, QR = 100 m and angle QAR = 30 degrees. We know, cot 30° = √3 = AQ/QR. Therefore, AQ = 100*√3 = 173 m.

Question 8

Amit is standing at a point P is watching the top of a tower, which makes an angle of elevation of 45° with Amit's eye. He walks some distance towards the tower to watch its top and the angle of elevation becomes 60°. What is the distance between the base of the tower and the point P?

• A. 4.2 units
• B. 8 units
• C. 10 units
• D. Data inadequate

Answer: (D)

Trigonometry & Height and Distances
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Let MN be the tower and Amit be standing at P (45° = angle MPN) and Q (60° = angle MQN). We are only given two angles and no sides of the triangles. Therefore, the data is inadequate.

Question 9

A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man's eye when at a distance of 60 meters from the tower. After 5 seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water?

• A. 32 kmph
• B. 36 kmph
• C. 40 kmph
• D. 44 kmph

Answer: (A)

Trigonometry & Height and Distances
Discuss it

Let the tower be PQ and the boat be at positions R and S when making angles of 45° and 30° respectively. Given, PR = 60 m. Now, PQ/PR = tan 45° = 1. So, PQ = PR = 60 m. Again, PQ/PS = tan 30° = 1/√3. So, PS = 60 * √3 m = 103.92 m. Distance covered in 5 seconds = 103.92 - 60 = 43.92 m. Speed in kmph = (43.92/5) * (18/5) = 32 kmph (approximately)

Question 10

What is the area of the circle which has the diagonal of the square as its diameter if the area of square is ' d ' ?

• A. πd
• B. πd²
• C. (1/4) πd²
• D. (1/2)πd

Answer: (D)

GATE CS 2018 General Aptitude Trigonometry & Height and Distances
Discuss it

One important observation to solve 
the question :

Diagonal of Square = Diameter of Circle.

Let side of square be x. 

From Pythagoras theorem.
Diagonal = √(2*x*x)

We know area of square = x * x = d

Diameter = Diagonal =  √(2*d)

Radius = √(d/2)
Area of Circle = π * √(d/2) * √(d/2) = 1/2 * π * d