Question

In the given figure, the angle of elevation of the top of a tower AC from a point B on the ground is $60^\circ$. If the height of the tower is 20m, find the distance of the point from the foot of the tower.

Hint:

In problems involving angles of elevation and right triangles, use trigonometric ratios like sine, cosine, and tangent to solve for unknown distances.

Answer 1

Step 1: Analyze the right triangle.
In this problem, we can form a right-angled triangle where the tower's height (AC) is the opposite side to the angle of elevation, and the distance from point B to the foot of the tower (AB) is the adjacent side. The angle of elevation is given as $60^\circ$.
Step 2: Apply trigonometric ratios.
We will use the tangent of the angle of elevation: \[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{AC}}{\text{AB}} \] Given that $\theta = 60^\circ$ and AC = 20m, we substitute the values into the equation: \[ \tan(60^\circ) = \frac{20}{\text{AB}} \]
Step 3: Solve for AB.
Since $\tan(60^\circ) = \sqrt{3}$, we have: \[ \sqrt{3} = \frac{20}{\text{AB}} \] Solving for AB: \[ \text{AB} = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3} \]
Step 4: Conclusion.
Thus, the distance from point B to the foot of the tower is: \[ \text{AB} = \frac{20\sqrt{3}}{3} \approx 11.55 \text{ m} \]