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Trigonometry & Height and Distances

Question 1
What is the maximum value of 5 Sinθ + 12 cosθ?
A
16
B
13
C
10
D
12
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Question 1 Explanation: 
Maximum value = √(a2 + b2) =13
Question 2
What is minimum value of Sinθ + cosθ ?
A
-2
B
1/2
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
2/√3
B
1/2
C
1/√2
D
1/√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 tan60o + tan120o?
A
1/2
B
0
C
2
D
2/3
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Question 4 Explanation: 
=tan(60) + tan(180-120) =tan(60)+(-tan(60)) {because in second quadrant tan is -ve} =0
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 √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. Given, AC:AB = √3 cot θ = √3 θ = 30 degrees.
Question 7
From a point P 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 P from the foot of the tower is:
A
149 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 QPR = 30 degrees. We know, cot 30° = √3 = PQ/QR. Therefore, PQ = 100*√3 = 173 m.
Question 8
A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30° with the man's eye. The man 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(3)^0.5 units
B
8 units
C
12 units
D
Data inadequate
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Question 8 Explanation: 
Let MN be the tower and man be standing at P (30° = 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)
There are 9 questions to complete.
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