Question:

If the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is $30^\circ$, then the height of the tower is :

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Remember the standard trigonometric values for common angles like $30^\circ, 45^\circ, 60^\circ$. For $30^\circ$, $\text{tan} 30^\circ = \frac{1}{\sqrt{3}}$. Always rationalize the denominator if your answer is in the form of a fraction with a radical in the denominator.
Updated On: Jun 5, 2025
  • $9\sqrt{3}$ m
  • $10\sqrt{3}$ m
  • $11\sqrt{3}$ m
  • $13\sqrt{3}$ m
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The Correct Option is B

Solution and Explanation

Step 1: Visualize the problem as a right-angled triangle.
Let the tower be represented by the vertical side (opposite side), and the distance from the foot of the tower to the observation point be the horizontal side (adjacent side). The angle of elevation is given. Step 2: Identify the given values.
Angle of elevation ($\theta$) = $30^\circ$
Distance from the foot of the tower (Adjacent side) = 30 m
Let the height of the tower be $h$ (Opposite side). Step 3: Choose the appropriate trigonometric ratio.
We have the angle, the adjacent side, and we need to find the opposite side. The tangent function relates these:
$\text{tan} \theta = \frac{\text{Opposite}}{\text{Adjacent}}$ Step 4: Substitute the given values and solve for the height. \[ \tan 30^\circ = \frac{h}{30} \] Recall that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), so: \[ \frac{1}{\sqrt{3}} = \frac{h}{30} \] Now, solve for \( h \): \[ h = \frac{30}{\sqrt{3}} \] To rationalize the denominator, multiply numerator and denominator by \( \sqrt{3} \): \[ h = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{30\sqrt{3}}{3} \] \[ h = 10\sqrt{3} \text{ m} \] Step 5: Final Answer.
The height of the tower is $10\sqrt{3}$ m. \[ \mathbf{(2)\ 10\sqrt{3}\ \text{m}} \]
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