Q11

1 markMCQSection A

A car is moving away from the base of a 30 m high tower. The angle of elevation of the top of the tower from the car at an instant, when the car is $10\sqrt{3}$ m away from the base of the tower, is : (a) 30° (b) 45° (c) 90° (d) 60°

Q: Which angle has tangent √3?

$\tan 60° = \sqrt{3}$. So the angle of elevation is 60°. (For reference, $\tan 30° = \frac{1}{\sqrt{3}}$ and $\tan 45° = 1$.)

Some Applications of Trigonometry

Angle of elevation

Options

(A)

30°

(B)

45°

(C)

90°

(D)

60°

Official Answer

(d) $60°$ — $\tan \theta = \frac{30}{10\sqrt{3}} = \sqrt{3}$ , so $\theta = 60°$ .

angle of elevationheights and distancestan θ = √360 degreestower height 30 mdistance 10√3

Marking Scheme

11 mark: correct option (d) $60°$ .
2Internal justification: $\tan \theta = \frac{30}{10\sqrt{3}} = \sqrt{3} \Rightarrow \theta = 60°$ .

Hint

Form $\tan \theta = \text{height}/\text{distance} = \frac{30}{10\sqrt{3}}$ , simplify to √3, and recall $\tan 60° = \sqrt{3}$ .

Quick Oral Answer

tan of the elevation equals height over distance, thirty over ten root three, which simplifies to root three, and since tan sixty degrees is root three, the angle is 60 degrees.

Analysis & Explanation

In the right triangle formed by the tower, the ground, and the line of sight, $\tan \theta = \frac{30}{10\sqrt{3}} = \sqrt{3}$ , so the angle of elevation $\theta = 60°$ .

Concept

Height-and-distance problem: tan θ = opposite/adjacent = height/horizontal distance.
Standard angle value: $\tan 60° = \sqrt{3}$ .

Key steps

$\tan \theta = \frac{30}{10\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$ (after rationalising).
$\theta = 60°$ → option (d).

Common mistakes

(a) 30°: $\tan 30° = \frac{1}{\sqrt{3}} \approx 0.577$ , which would need a much larger distance than height — doesn't fit here.
(b) 45°: $\tan 45° = 1$ needs distance = height, which isn't the case ( $30 \ne 10\sqrt{3}$ ).
(c) 90°: impossible, since that would require zero horizontal distance (car directly under the top).

real-world

As a car moves further from a tower, the angle of elevation keeps decreasing — here the relatively small distance (10√3 m against a 30 m tower) correctly gives a fairly steep 60°.

Common Mistakes

1Leaving $\frac{30}{10\sqrt{3}}$ as $\frac{3}{\sqrt{3}}$ without recognising it equals $\sqrt{3}$ , then guessing the angle.
2Inverting the ratio to distance/height and getting $\tan \theta = \frac{1}{\sqrt{3}}$ , leading to the wrong answer $30°$ .
3Forgetting to rationalise the surd and mis-simplifying $\frac{3}{\sqrt{3}}$ .

Interesting Facts

The relationship 'as you move away, the angle of elevation shrinks' is exactly how surveyors and navigators estimate the height of hills and buildings from a distance.

$\tan 60° = \sqrt{3} \approx 1.732$ is one of the standard exact values; 60° elevation is quite steep — you would be tilting your head well back to see the top.

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Frequently Asked Questions

How do I simplify 30/(10√3)?

Divide numerator and denominator by 10 to get 3/√3, then rationalise: $\frac{3}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}$ . So $\tan \theta = \sqrt{3}$ .

Which angle has tangent √3?

$\tan 60° = \sqrt{3}$ . So the angle of elevation is 60°. (For reference, $\tan 30° = \frac{1}{\sqrt{3}}$ and $\tan 45° = 1$ .)

Why can't the angle be 90°?

A $90°$ elevation means looking straight up, which happens only when the car is right at the base (distance 0). Here the car is $10\sqrt{3}$ m away, so the angle must be less than $90°$ .