Question 1: The height of a tree is \sqrt{3} times the length of its shadow. Find the angle of elevation of the sun.

Answer:

Let the length of the shadow = x

Therefore the height of the tree = \sqrt{3} x

Therefore \tan \  \theta = \frac{\sqrt{3} x}{x} = \sqrt{3} \Rightarrow \theta = 60^o

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Question 2: The angle of elevation of the top of a tower, from a point on the ground and at a distance of 160 \ m from its foot, is found to be 60^o .Find the height of the tower.

Answer:

Let the height of the tower = x

Therefore tan \ 60^o = \frac{x}{160} \Rightarrow x = 160 \times tan \ 60^o = 277.13 \ m

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Question 3: A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 \ m from the wall and the ladder is making an angle of 68^o with the ground. Find the height, up to which the ladder reaches.

Answer:

Let the height to which the ladder reaches = h

Distance from the base of the wall = 2.4 \ m

Therefore \frac{h}{2.4} = tan \ 68^o

\Rightarrow h = 2.4 \times 2.475 = 5.94 \ m

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Question 4: Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30^o and 38^o respectively. Find the distance between them, if the height of the tower is 50 \ m .

Answer:

Let the distance of the first person from the tower = x_1

Let the distance of the second person from the tower = x_2

Therefore \frac{50}{x_1} = tan \ 30^o \Rightarrow x_1 = \frac{50}{tan \ 30^o} = 86.60 \ m

Similarly \frac{50}{x_2} = tan \ 38^o \Rightarrow x_2 = \frac{50}{tan \ 38^o} = 64.00 \ m

Therefore the distance between the two persons = x_1 + x_2 = 86.60+64 = 150.6 \ m

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Question 5: A kite is attached to a sling. Find the length of the string, when the height of the kite is 60 \ m and the string make an angle 30^o with the ground.

Answer:

Height of the kite = 60 \ m

Let the length of the string = x 

Therefore \frac{60}{x} = sin \ 30 \Rightarrow x = 60 \times 2 = 120 \ m

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Question 6: A boy, 1.6 \ m tall, is 20 \ m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45^o (ii) 60^o . Find the height of the tower in each case.

Answer:

Case 1: Angle of elevation = 45^o

Distance of the boy from the tower = 20 \ m

Let the height of the tower = h_1 + 1.6

Now \frac{h_1}{20} = tan \ 45^o \Rightarrow h_1 = 20 \ m

Hence the height of the tower = 20 + 1.6 = 21.6

Case 2: Angle of elevation = 60^o

Distance of the boy from the tower = 20 \ m

Let the height of the tower = h_2 + 1.6

Now \frac{h_2}{20} = tan \ 60^o \Rightarrow h_2 = 20\sqrt{3}  = 34.64 \ m

Hence the height of the tower = 34.64 + 1.6 = 36.24 \ m

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Question 7: The upper part of a tree, broken over by the wind, makes an angle of 45^o with the ground; and the distance from the root to the point where the top of the tree touches the ground, is 15 \ m . What was the height of the tree before it was broken?

Answer:

Let the broken part of the tree is h_2 and the part of the tree still standing upright be h_1

Therefore \frac{h_1}{15} = tan \ 45^o \Rightarrow h_1 = 15 \ m

Similarly, \frac{15}{h_2} = sin \ 45^o \Rightarrow h_2 = 15 \times \sqrt{2} = 21.21 \ m

Hence the height of the tree before it was broken = 15 + 21.21 = 36.21 \ m

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Question 8: The angle of elevation of the top of an unfinished tower at a point distance 80 \ m from its base is 30^o . How much higher must the tower be raised so that its angle of elevation at the same point may be 60^o ?

Answer:

Let the height of the unfinished structure = h

Therefore  \frac{h}{80} = tan \ 30^o \Rightarrow h = 80 \times tan\ 30^o = 46.19 \ m

Let the tower be raised  = x \ m

Therefore \frac{x+46.16}{80} = tan \ 60^o \Rightarrow x = 80 tan \ 60^o - 46.19 = 92.37 \ m

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Question 9: At a particular time when sun’s altitude is 30^o , the length of the shadow of altitude is a vertical tower is 45 \ m . Calculate:

(i) the height of the tower,

(ii) the length of the shadow of the tower, when the sun’s altitude is same 45^o (b) 60^o .

Answer:

Let the height of the tower = h

(i) Therefore \frac{h}{45} = tan \ 30^o \Rightarrow h = 45 \times tan \ 30^o = 25.98 \ m

(ii) Let the shadow be x_1 when sun’s altitude is same 45^o and x_2 when sun’s altitude is 60^o .

Therefore \frac{25.98}{x_1} = tan \ 45^o \Rightarrow x_1 = 25.98 \ m

Similarly, \frac{25.98}{x_2} = tan \ 60^o \Rightarrow x_1 = \frac{25.98}{tan \ 60^o} = 15 \ m

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Question 10: The vertical poles are on either side of a road. A 30 \ m long ladder is placed between the two poles. When the ladder rests against one pole, it makes angle 32^o 24' with the pole and when it is turned to rest against another pole, it makes angle 32^o 24' with the road. Calculate the width of the road.

Answer:

Let the distance of the foot of the ladder from the towers be x_1 and x_2

For first tower, the angle of elevation = 90- 32.4 = 57.6^o

Therefore \frac{x_1}{30} = cos \ 57.6^o \Rightarrow x_1 = 16.07 \ m

For first tower, the angle of elevation = 32.4^o

Therefore \frac{x_2}{30} = cos \ 32.4^o \Rightarrow x_2 = 25.32 \ m

Therefore the width of the road = 16.07 + 25.32 = 41.4 \ m

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Question 11: Two climbers are at points A and B on a vertical cliff face. To an observer C , 40 \ m from the foot of the cliff, on the level ground, A is at an elevation of 48^o and B of 57^o . What is the distance between the climbers?

Answer:

Let the climber B be at height h_1 and climber A be at h_2 height

Therefore \frac{h_1}{40} = tan \ 57^o \Rightarrow h_1 = 40 \times tan \ 57^o

Similarly \frac{h_2}{40} = tan \ 48^o \Rightarrow h_2 = 40 \times tan \ 48^o

Therefore the distance between the two climbers = h_1 - h_2 = 40 \times tan \ 57^o - 40 \times tan \ 48^o = 40 \times 0.42925 = 17.17 \ m 

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Question 12: A man stands 9 \ m away from a flag-pole. He observes that angle of elevation of the top of the pole is 28^o and the angle of depression of the bottom of the pole is 13^o . Calculate the height of the pole.

Answer:

Let the height of the top of the pole = h_1

Let the height of the bottom of the pole = h_2

Therefore \frac{h_1}{9} = tan \ 28^o

The angle of elevation of the bottom of the pole = angle of elevation of the bottom of the pole = 13^o

Therefore \frac{h_2}{9} = tan \ 13^o

Hence the height of the pole = h_1 - h_2 = 9 \ tan \ 28^o - 9 \ tan \ 13^o = 2.71 \ m

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Question 13: From the top of a cliff 92 \ m height, the angle of depression of a buoy is 20^o . Calculate, to the nearest meter, the distance of the buoy from the foot of the cliff.

Answer:

Angle of depression = angle of elevation = 20^o

Let the distance of the buoy from the foot of the cliff = x

Therefore x = \frac{92}{tan \ 20^o} = 252.77 \approx 253 \ m

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