Question

The angle of elevation of a 15 m high tower from a point 15 m away from its foot is:

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

Hint:

When the height of an object and the horizontal distance from its base are equal, the angle of elevation to its top will always be $45^\circ$, because $\text{tan} 45^\circ = 1$. This creates an isosceles right-angled triangle.

Answer 1

The Correct Option is B

Step 1: Visualize the problem as a right-angled triangle.
Let the tower be represented by the vertical side (opposite side) of a right-angled triangle, and the distance from the foot of the tower to the observation point be the horizontal side (adjacent side). The angle of elevation is the angle between the horizontal line and the line of sight to the top of the tower. Step 2: Identify the given values.
Height of the tower (Opposite side) = 15 m
Distance from the foot of the tower (Adjacent side) = 15 m Step 3: Choose the appropriate trigonometric ratio.
We have the opposite side and the adjacent side, so the tangent function is suitable:
\[ \tan(\text{angle of elevation}) = \frac{\text{Opposite}}{\text{Adjacent}} \] Step 4: Substitute the given values and solve for the angle. Let the angle of elevation be \( \theta \). \[ \tan \theta = \frac{15 \text{ m}}{15 \text{ m}} \] \[ \tan \theta = 1 \] We know that \( \tan(45^\circ) = 1 \). Therefore, \( \theta = 45^\circ \). Step 5: Final Answer. The angle of elevation of the tower is $45^\circ$. \[ \mathbf{(2)\ 45^\circ} \]