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: Start from the cotangent ratio instead of the tangent ratio.
Instead of writing $\tan \theta = \frac{\text{height}}{\text{distance}}$ straight away, let's set up the ratio the other way round first, using base over perpendicular, and then flip it at the end.

Step 2: Set up the triangle.
Let the tower be $AB$ with height $AB = 30\text{ m}$, and let $C$ be the car's position with $BC = 10\sqrt{3}\text{ m}$. The angle of elevation from the car to the top of the tower is $\theta = \angle ACB$.

Step 3: Write the cotangent ratio.
\[ \cot \theta = \frac{\text{base}}{\text{perpendicular}} = \frac{BC}{AB} = \frac{10\sqrt{3}}{30} \]
Simplify the fraction:
\[ \cot \theta = \frac{10\sqrt{3}}{30} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \]

Step 4: Flip to get the tangent value and read off the angle.
Since $\cot \theta = \frac{1}{\sqrt{3}}$, taking the reciprocal gives:
\[ \tan \theta = \sqrt{3} \]
From the standard trigonometric table, $\tan 60^\circ = \sqrt{3}$, so:
\[ \theta = 60^\circ \]

Step 5: Final answer.
The angle of elevation is $60^\circ$, which is option (D).
\[ \boxed{\theta = 60^\circ} \]