# Trigonometry & Height and Distances

• Last Updated : 15 May, 2016

 Question 1

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

 A 12 B 5 C 6 D 1
Trigonometry & Height and Distances
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Question 1 Explanation:

Maximum value = √(a2 + b2) = 5

 Question 2

What is minimum value of Sinθ + cosθ ?

 A -2 B √3/2 C -1 D -√2
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Question 2 Explanation:

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
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Question 3 Explanation:

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

 Question 4

Find the Value of tan30o + tan120o?

 A √3 B -1/√3 C 0 D -2/√3
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Question 4 Explanation:

=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
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Question 5 Explanation:
= 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
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Question 6 Explanation:

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
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Question 7 Explanation:

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
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Question 8 Explanation:

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
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Question 9 Explanation:
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
GATE CS 2018    General Aptitude    Trigonometry & Height and Distances
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Question 10 Explanation: ```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)