Question

A car is moving away from the base of a $30\text{ m}$ high tower. The angle of elevation of the top of the tower from the car at an instant, when the car is $10\sqrt{3}\text{ m}$ away from the base of the tower, is :

• A. $30^\circ$
• B. $45^\circ$
• C. $90^\circ$
• D. $60^\circ$

Hint:

Always remember the standard ratios of a $30^\circ-60^\circ-90^\circ$ triangle.
In such a triangle, the ratio of the side opposite to $30^\circ$, the side opposite to $60^\circ$, and the hypotenuse is $1 : \sqrt{3} : 2$.
Here, since the height ($30\text{ m}$) is $\sqrt{3}$ times the base ($10\sqrt{3}\text{ m}$), the angle opposite to the height must be $60^\circ$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The topic of this question is Heights and Distances, which is an application of basic trigonometry.
We are given a vertical tower and a car on the ground moving away from its base.
At a specific instant, we know the height of the tower and the horizontal distance of the car from the base of the tower.
We need to determine the angle of elevation of the top of the tower from the position of the car at this instant.

Step 2: Key Formula or Approach:
We can model this scenario using a right-angled triangle where:

• The perpendicular represents the height of the tower ($h$).

• The base represents the horizontal distance of the car from the tower's base ($d$).

• The angle of elevation is denoted by $\theta$.

The trigonometric ratio that links the perpendicular and the base of a right-angled triangle is the tangent function:
\[ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \]

Step 3: Detailed Explanation:

• Let $AB$ represent the vertical tower of height $h = 30\text{ m}$.

• Let $C$ represent the position of the car on the ground.

• The distance from the base of the tower $B$ to the car $C$ is given as $BC = 10\sqrt{3}\text{ m}$.

• Let $\theta$ be the angle of elevation of the top of the tower $A$ from the car $C$, which is $\angle ACB$.

• In the right-angled triangle $ABC$, apply the tangent ratio:
\[ \tan \theta = \frac{AB}{BC} \]

• Substitute the given values into the formula:
\[ \tan \theta = \frac{30}{10\sqrt{3}} \]

• Simplify the fraction:
\[ \tan \theta = \frac{3}{\sqrt{3}} \]

• Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$:
\[ \tan \theta = \frac{3\sqrt{3}}{3} = \sqrt{3} \]

• We know from standard trigonometric values that:
\[ \tan 60^\circ = \sqrt{3} \]

• Therefore, equating the values gives:
\[ \theta = 60^\circ \]


Step 4: Final Answer:
The angle of elevation of the top of the tower from the car is $60^\circ$, which corresponds to option (D).