Question

From a point on the ground, which is 60 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is found to be \(45^\circ\). The height (in metres) of the tower is :

• A. \(10\sqrt{3}\)
• B. \(30\sqrt{3}\)
• C. 60
• D. 30

Hint:

In any right-angled triangle, if one of the acute angles is \(45^\circ\), the triangle is isosceles because the other acute angle is also \(180^\circ - 90^\circ - 45^\circ = 45^\circ\).
Therefore, the perpendicular and the base are always equal in length.
Since the distance to the base is 60 m, the height must automatically be 60 m!

Answer 1

The Correct Option is C

Step 1: Spot the isosceles right triangle instead of computing with the tangent ratio.
Let $AB$ be the tower of height $h$ and $C$ the point on the ground, 60 m from the foot $B$, with $\angle ACB = 45^\circ$ the angle of elevation to the top $A$.
Triangle $ABC$ is right angled at $B$ (the tower stands vertical to the ground). Since one of its other two angles is $\angle ACB = 45^\circ$, the third angle must be:
\[ \angle BAC = 180^\circ - 90^\circ - 45^\circ = 45^\circ \]

Step 2: Recognize what two equal angles mean for the sides.
Since $\angle ACB = \angle BAC = 45^\circ$, triangle $ABC$ has two equal angles, so the sides opposite these equal angles must also be equal (equal angles sit opposite equal sides).
The side opposite $\angle ACB$ is $AB$ (the tower height $h$), and the side opposite $\angle BAC$ is $BC$ (the base distance, 60 m).

Step 3: Equate the two sides.
\[ AB = BC \]
\[ h = 60 \text{ m} \]

Step 4: Final Answer.
The height of the tower is 60 metres, so option (C) is correct. \[ \boxed{60 \text{ m}} \]