The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Step-by-Step Solution
Key Concept: Use the definition of the tangent function in a right‑angled triangle: \(\tan \theta = \dfrac{\text{opposite side}}{\text{adjacent side}}\). Here the height of the tower is the side opposite the angle of elevation and the given distance of 30 m is the adjacent side.
1. Draw a diagram – Let \(AB\) be the tower with \(A\) at the foot and \(B\) at the top. Let \(C\) be the point on the ground 30 m from the foot. Then \(\angle ACB = 30^{\circ}\) (angle of elevation) and \(AC = 30\) m.
2. Identify the right‑angled triangle – \(\triangle ABC\) is a right‑angled triangle with \(\angle BAC = 90^{\circ}\). The side \(AB\) (height of the tower) is opposite the angle \(30^{\circ}\), and \(AC\) is the adjacent side.
3. Apply the tangent definition:
$$\tan 30^{\circ} = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{AC}$$
4. Substitute the known values:
$$\tan 30^{\circ} = \frac{AB}{30}\
\Rightarrow AB = 30 \times \tan 30^{\circ}$$
5. Use the exact value \(\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\):
$$AB = 30 \times \frac{\sqrt{3}}{3} = 10\sqrt{3}\text{ m}$$
6. If a decimal answer is required, evaluate \(10\sqrt{3} \approx 10 \times 1.732 = 17.32\) m.
7. State the result – The height of the tower is \(10\sqrt{3}\) m (approximately \(17.3\) m).
Correct Answer: 10√3 m (≈ 17.3 m)