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Question 1: A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. The height of the tower is:
Solution: (A) $15\sqrt{3}$ m
Step 1: Let height be $h$. $\tan 60^\circ = \frac{h}{15}$.
Step 2: $\sqrt{3} = \frac{h}{15} \Rightarrow h = 15\sqrt{3}$ m. -
Question 2: The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30°. The distance of the car from the base of the tower is:
Solution: (C) $75\sqrt{3}$ m
Step 1: Angle of elevation from car to tower top is also 30°. Let distance be $d$.
Step 2: $\tan 30^\circ = \frac{75}{d} \Rightarrow \frac{1}{\sqrt{3}} = \frac{75}{d} \Rightarrow d = 75\sqrt{3}$ m. -
Question 3: A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall is:
Solution: (C) 7.5 m
Step 1: Angle with wall is 60°, so angle with ground is $90^\circ - 60^\circ = 30^\circ$.
Step 2: Let height be $h$. $\sin 30^\circ = \frac{h}{15}$.
Step 3: $\frac{1}{2} = \frac{h}{15} \Rightarrow h = 7.5$ m. -
Question 4: A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. The length of the string is:
Solution: (A) $40\sqrt{3}$ m
Step 1: Let length be $L$. $\sin 60^\circ = \frac{60}{L}$.
Step 2: $\frac{\sqrt{3}}{2} = \frac{60}{L} \Rightarrow L = \frac{120}{\sqrt{3}} = 40\sqrt{3}$ m. -
Question 5: If the shadow of a tower is equal to its height, then the angle of elevation of the sun is:
Solution: (B) 45°
Step 1: Let height be $h$. Shadow is also $h$.
Step 2: $\tan \theta = \frac{\text{height}}{\text{shadow}} = \frac{h}{h} = 1$.
Step 3: $\theta = 45^\circ$. -
Question 6: A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. The height of the tree is:
Solution: (A) $8\sqrt{3}$ m
Step 1: Standing part $y = 8 \tan 30^\circ = \frac{8}{\sqrt{3}}$.
Step 2: Broken part $x = \frac{8}{\cos 30^\circ} = \frac{16}{\sqrt{3}}$.
Step 3: Total height $= x+y = \frac{24}{\sqrt{3}} = 8\sqrt{3}$ m. -
Question 7: The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. The height of the tower is:
Solution: (A) 6 m
Step 1: Let height be $h$. $\tan \theta = h/9$ and $\tan(90-\theta) = h/4$.
Step 2: $\cot \theta = h/4$.
Step 3: $\tan \theta \cdot \cot \theta = 1 \Rightarrow (\frac{h}{9})(\frac{h}{4}) = 1 \Rightarrow h^2 = 36 \Rightarrow h = 6$ m. -
Question 8: If the height of a pole is $2\sqrt{3}$ meters and the length of its shadow is 2 meters, find the angle of elevation of the sun.
Solution: (C) 60°
Step 1: $\tan \theta = \frac{\text{height}}{\text{shadow}} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Step 2: $\theta = 60^\circ$. -
Question 9: The angle formed by the line of sight with the horizontal when the object viewed is above the horizontal level is called:
Solution: (B) Angle of elevation
Reason: This is the definition of the angle of elevation. -
Question 10: From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 3 m from the banks, the width of the river is:
Solution: (B) $3(\sqrt{3}+1)$ m
Step 1: Let width parts be $x_1, x_2$.
Step 2: $\tan 30^\circ = 3/x_1 \Rightarrow x_1 = 3\sqrt{3}$.
Step 3: $\tan 45^\circ = 3/x_2 \Rightarrow x_2 = 3$.
Step 4: Total width $= x_1 + x_2 = 3\sqrt{3} + 3 = 3(\sqrt{3}+1)$ m.