NCERT Class X Chapter 9: Some Application Of Trigonometry Exercise 9.1 Question 8

NCERT Class X Chapter 9: Some Applications of Trigonometry

Question:

A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Given:

  • Height of the statue = 1.6 m
  • Angle of elevation of the top of the statue = 60°
  • Angle of elevation of the top of the pedestal = 45°

To Find:

Height of the pedestal

Formula:

\[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \]

Solution:

Step 1: Let the height of the pedestal be \( h \) meters. The total height of the statue and pedestal is \( h + 1.6 \) meters. Let the distance from the point on the ground to the base of the pedestal be \( x \) meters.

Step 2: Using the angle of elevation to the top of the pedestal (45°):

\[ \tan 45^\circ = \frac{h}{x} \] \[ 1 = \frac{h}{x} \implies h = x \]

Step 3: Using the angle of elevation to the top of the statue (60°):

\[ \tan 60^\circ = \frac{h + 1.6}{x} \] \[ \sqrt{3} = \frac{h + 1.6}{x} \]

Step 4: Substitute \( h = x \) from Step 2 into the equation from Step 3:

\[ \sqrt{3} = \frac{h + 1.6}{h} \] \[ \sqrt{3} = 1 + \frac{1.6}{h} \] \[ \sqrt{3} - 1 = \frac{1.6}{h} \] \[ h = \frac{1.6}{\sqrt{3} - 1} \]

Step 5: Calculate the value of \( h \):

\[ \sqrt{3} \approx 1.732 \] \[ h = \frac{1.6}{1.732 - 1} \] \[ h = \frac{1.6}{0.732} \] \[ h \approx 2.18 \ \text{m} \]

Result:

The height of the pedestal is approximately \( 2.18 \) meters.

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