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

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.

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.

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 $50 \ m$.

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.

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.

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?

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 $60^o$?

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) $60^o$.

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.

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?

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.

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.

Angle of depression = angle of elevation $= 20^o$
Let the distance of the buoy from the foot of the cliff $= x$
Therefore $x =$ $\frac{92}{tan \ 20^o}$ $= 252.77 \approx 253 \ m$