NCERT Class X Chapter 9: Some Application Of Trigonometry Example 5

NCERT Class X Chapter 9: Some Applications of Trigonometry

Question:

The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.

Given:

Difference in shadow lengths = 40 m
Sun's altitude (angle of elevation) = 30° and 60°

To Find:

Height of the tower

Formula:

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

Solution:

Step 1: Let the height of the tower be \( h \) meters. Let the length of the shadow when the Sun's altitude is 60° be \( x \) meters.

Step 2: When the Sun's altitude is 60°, using the definition of tangent:

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

Step 3: When the Sun's altitude is 30°, the length of the shadow is \( x + 40 \) meters.

\[ \tan 30^\circ = \frac{h}{x + 40} \] \[ \frac{1}{\sqrt{3}} = \frac{h}{x + 40} \] \[ x + 40 = h \sqrt{3} \]

Step 4: Substitute the value of \( x \) from Step 2 into the equation from Step 3:

\[ x + 40 = h \sqrt{3} \] \[ \frac{h}{\sqrt{3}} + 40 = h \sqrt{3} \] \[ 40 = h \sqrt{3} - \frac{h}{\sqrt{3}} \] \[ 40 = \frac{3h - h}{\sqrt{3}} \] \[ 40 = \frac{2h}{\sqrt{3}} \]

Step 5: Solve for \( h \):

\[ 2h = 40\sqrt{3} \] \[ h = \frac{40\sqrt{3}}{2} \] \[ h = 20\sqrt{3} \]

Result:

The height of the tower is \( 20\sqrt{3} \) m.

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