NCERT Class X Chapter 9: Some Application Of Trigonometry Exercise 9.1 Question 9
NCERT Class X Chapter 9: Some Applications of Trigonometry
Question:
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Given:
Angle of elevation of the top of the building from the foot of the tower = \(30^\circ\)
Angle of elevation of the top of the tower from the foot of the building = \(60^\circ\)
Height of the tower = \(50\,\text{m}\)
To Find:
Height of the building.
Formula:
Trigonometric ratio:
\[
\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}
\]
Solution:
Step 1: Let the height of the building be \(h\) meters. Let the distance between the building and the tower be \(x\) meters.
Step 2: From the foot of the tower, the angle of elevation to the top of the building is \(30^\circ\):
\[ \tan 30^\circ = \frac{h}{x} \] \[ \frac{1}{\sqrt{3}} = \frac{h}{x} \] \[ x = h \sqrt{3} \]Step 3: From the foot of the building, the angle of elevation to the top of the tower is \(60^\circ\):
\[ \tan 60^\circ = \frac{50}{x} \] \[ \sqrt{3} = \frac{50}{x} \] \[ x = \frac{50}{\sqrt{3}} \]Step 4: Equate the two values of \(x\) obtained in Steps 2 and 3:
\[ h \sqrt{3} = \frac{50}{\sqrt{3}} \] \[ h = \frac{50}{\sqrt{3} \times \sqrt{3}} \] \[ h = \frac{50}{3} \] \[ h \approx 16.67 \text{ m} \]Step 5: There is a calculation error in the previous step. Let's carefully solve for \(h\) by setting the two expressions for \(x\) equal:
\[ h \sqrt{3} = \frac{50}{\sqrt{3}} \] \[ h = \frac{50}{\sqrt{3} \times \sqrt{3}} \] \[ h = \frac{50}{3} \] \[ h \approx 16.67 \text{ m} \]Result:
The height of the building is approximately \(16.67\,\text{m}\).
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