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

Answer:

Let the length of the shadow \displaystyle = x  

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

\displaystyle \text{Therefore } \tan \theta = \frac{\sqrt{3} x}{x} = \sqrt{3} \Rightarrow \theta = 60^{\circ}  

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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 \displaystyle 160 \text{ m } from its foot, is found to be \displaystyle 60^{\circ} .Find the height of the tower.

Answer:

Let the height of the tower \displaystyle = x  

\displaystyle \text{Therefore } \tan 60^{\circ} = \frac{x}{160} \Rightarrow x = 160 \times \tan 60^{\circ} = 277.13 \text{ 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 \displaystyle 2.4 \text{ m } from the wall and the ladder is making an angle of \displaystyle 68^{\circ} with the ground. Find the height, up to which the ladder reaches.

Answer:

Let the height to which the ladder reaches \displaystyle = h  

Distance from the base of the wall \displaystyle = 2.4 \text{ m }  

\displaystyle \text{Therefore } \frac{h}{2.4} = \tan 68^{\circ}  

\displaystyle \Rightarrow h = 2.4 \times 2.475 = 5.94 \text{ 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 \displaystyle 30^{\circ} and \displaystyle 38^{\circ} respectively. Find the distance between them, if the height of the tower is \displaystyle 50 \text{ m } .

Answer:

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

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

\displaystyle \text{Therefore } \frac{50}{x_1} = \tan 30^{\circ} \Rightarrow x_1 = \frac{50}{\tan 30^{\circ} } = 86.60 \text{ m }  

\displaystyle \text{Similarly } \frac{50}{x_2} = \tan 38^{\circ} \Rightarrow x_2 = \frac{50}{\tan 38^{\circ} } = 64.00 \text{ m }  

Therefore the distance between the two persons \displaystyle = x_1 + x_2 = 86.60+64 = 150.6 \text{ 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 \displaystyle 60 \text{ m } and the string make an angle \displaystyle 30^{\circ} with the ground.

Answer:

Height of the kite \displaystyle = 60 \text{ m }  

Let the length of the string \displaystyle = x  

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

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

Answer:

Case 1: Angle of elevation \displaystyle = 45^{\circ}  

Distance of the boy from the tower \displaystyle = 20 \text{ m }  

Let the height of the tower \displaystyle = h_1 + 1.6  

\displaystyle \text{Now } \frac{h_1}{20} = \tan 45^{\circ} \Rightarrow h_1 = 20 \text{ m }  

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

Case 2: Angle of elevation \displaystyle = 60^{\circ}  

Distance of the boy from the tower \displaystyle = 20 \text{ m }  

Let the height of the tower \displaystyle = h_2 + 1.6  

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

Hence the height of the tower \displaystyle = 34.64 + 1.6 = 36.24 \text{ m }  

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

Answer:

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

\displaystyle \text{Therefore } \frac{h_1}{15} = \tan 45^{\circ} \Rightarrow h_1 = 15 \text{ m }  

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

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

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

Answer:

Let the height of the unfinished structure \displaystyle = h  

\displaystyle \text{Therefore } \frac{h}{80} = \tan 30^{\circ} \Rightarrow h = 80 \times \tan 30^{\circ} = 46.19 \text{ m }  

Let the tower be raised \displaystyle = x \text{ m }  

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

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

(i) the height of the tower,

(ii) the length of the shadow of the tower, when the sun’s altitude is same \displaystyle 45^{\circ} (b) \displaystyle 60^{\circ} .

Answer:

Let the height of the tower \displaystyle = h  

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

(ii) Let the shadow be \displaystyle x_1 when sun’s altitude is same \displaystyle 45^{\circ} and \displaystyle x_2 when sun’s altitude is \displaystyle 60^{\circ} .

\displaystyle \text{Therefore } \frac{25.98}{x_1} = \tan 45^{\circ} \Rightarrow x_1 = 25.98 \text{ m }  

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

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

Answer:

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

For first tower, the angle of elevation \displaystyle = 90- 32.4 = 57.6^{\circ}  

\displaystyle \text{Therefore } \frac{x_1}{30} = \cos 57.6^{\circ} \Rightarrow x_1 = 16.07 \text{ m }  

For first tower, the angle of elevation \displaystyle = 32.4^{\circ}  

\displaystyle \text{Therefore } \frac{x_2}{30} = \cos 32.4^{\circ} \Rightarrow x_2 = 25.32 \text{ m }  

Therefore the width of the road \displaystyle = 16.07 + 25.32 = 41.4 \text{ m }  

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

Answer:

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

\displaystyle \text{Therefore } \frac{h_1}{40} = \tan 57^{\circ} \Rightarrow h_1 = 40 \times \tan 57^{\circ}  

\displaystyle \text{Similarly } \frac{h_2}{40} = \tan 48^{\circ} \Rightarrow h_2 = 40 \times \tan 48^{\circ}  

Therefore the distance between the two climbers \displaystyle = h_1 - h_2 = 40 \times \tan 57^{\circ} - 40 \times \tan 48^{\circ} = 40 \times 0.42925 = 17.17 \text{ m }  

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

Answer:

Let the height of the top of the pole \displaystyle = h_1  

Let the height of the bottom of the pole \displaystyle = h_2  

\displaystyle \text{Therefore } \frac{h_1}{9} = \tan 28^{\circ}  

The angle of elevation of the bottom of the pole = angle of elevation of the bottom of the pole \displaystyle = 13^{\circ}  

\displaystyle \text{Therefore } \frac{h_2}{9} = \tan 13^{\circ}  

Hence the height of the pole \displaystyle = h_1 - h_2 = 9 \tan 28^{\circ} - 9 \tan 13^{\circ} = 2.71 \text{ m }  

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

Answer:

Angle of depression = angle of elevation \displaystyle = 20^{\circ}  

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

\displaystyle \text{Therefore } x = \frac{92}{\tan 20^{\circ} } = 252.77 \approx 253 \text{ m }