Question

The angle of elevation of the top of a hill at the foot of the tower is \(60^\circ\) and the angle of elevation of the top of the tower from the foot of the hill is \(30^\circ\). If the tower is \(50\) m high, what is the height of the hill?

• A. \(180\) m
• B. \(120\) m
• C. \(100\) m
• D. \(150\) m

Hint:

Remember:

• \[ \tan\theta = \frac{\text{Height}}{\text{Distance}} \]
• \[ \tan60^\circ = \sqrt{3} \]
• \[ \tan30^\circ = \frac{1}{\sqrt{3}} \]

Answer 1

The Correct Option is D

Concept: For problems involving angles of elevation, use: \[ \tan\theta = \frac{\text{Perpendicular}}{\text{Base}} \] If two objects are observed from each other’s foot, the horizontal distance between them remains the same.

Step 1: Assume the height of the hill and horizontal distance. Let: \[ \text{Height of hill} = h \] Let the horizontal distance between the tower and hill be: \[ x \] Height of tower: \[ 50\,\text{m} \]

Step 2: Use angle of elevation from foot of tower to top of hill. Given: \[ \theta = 60^\circ \] Using: \[ \tan60^\circ = \frac{h}{x} \] \[ \sqrt{3} = \frac{h}{x} \] \[ h = \sqrt{3}x \quad \cdots (1) \]

Step 3: Use angle of elevation from foot of hill to top of tower. Given: \[ \theta = 30^\circ \] Using: \[ \tan30^\circ = \frac{50}{x} \] \[ \frac{1}{\sqrt{3}} = \frac{50}{x} \] \[ x = 50\sqrt{3} \]

Step 4: Substitute value of \(x\) into equation (1). \[ h = \sqrt{3}(50\sqrt{3}) \] \[ h = 50 \times 3 \] \[ h = 150\,\text{m} \] Therefore, \[ \boxed{150\,\text{m}} \]