NCERT Class X Chapter 9: Some Application Of Trigonometry Example 6
NCERT Class X Chapter 9: Some Applications of Trigonometry Example 6
Question:
The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multi-storeyed building and the distance between the two buildings.
Given:
- Angle of depression of the top of the building = \(30^\circ\)
- Angle of depression of the bottom of the building = \(45^\circ\)
- Height of the smaller building = \(8\,\text{m}\)
To Find:
- Height of the multi-storeyed building
- Distance between the two buildings
Formula:
\[ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \]
Solution:
Step 1: Let the height of the multi-storeyed building be \(h\) meters and the distance between the two buildings be \(x\) meters.
Step 2: Consider the angle of depression to the bottom of the 8 m building (\(45^\circ\)).
From the top of the multi-storeyed building, the vertical distance to the ground is \(h\), and the horizontal distance is \(x\).
\[ \tan 45^\circ = \frac{h}{x} \]Since \(\tan 45^\circ = 1\):
\[ 1 = \frac{h}{x} \implies h = x \]Step 3: Consider the angle of depression to the top of the 8 m building (\(30^\circ\)).
The vertical distance from the top of the multi-storeyed building to the top of the 8 m building is \(h - 8\), and the horizontal distance remains \(x\).
\[ \tan 30^\circ = \frac{h - 8}{x} \]Since \(\tan 30^\circ = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{h - 8}{x} \]Step 4: Substitute \(h = x\) from Step 2 into the equation from Step 3 and solve for \(x\).
\[ \frac{1}{\sqrt{3}} = \frac{x - 8}{x} \]Cross-multiplying:
\[ x - 8 = \frac{x}{\sqrt{3}} \] \[ x - \frac{x}{\sqrt{3}} = 8 \] \[ x \left(1 - \frac{1}{\sqrt{3}}\right) = 8 \] \[ x = \frac{8}{1 - \frac{1}{\sqrt{3}}} \]Rationalizing the denominator:
\[ x = \frac{8}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = 8 \times \frac{\sqrt{3}}{\sqrt{3} - 1} \]Multiply numerator and denominator by \((\sqrt{3} + 1)\):
\[ x = 8 \times \frac{\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \] \[ x = 8 \times \frac{3 + \sqrt{3}}{3 - 1} \] \[ x = 8 \times \frac{3 + \sqrt{3}}{2} \] \[ x = 4(3 + \sqrt{3}) \]Calculate the value:
\[ x = 4(3 + 1.732) = 4 \times 4.732 = 18.928 \]So, \(x \approx 18.93\,\text{m}\).
Step 5: The height of the multi-storeyed building is \(h = x\).
\[ h \approx 18.93\,\text{m} \]Result:
Height of the multi-storeyed building \( \approx 18.93\,\text{m} \)
Distance between the two buildings \( \approx 18.93\,\text{m} \)
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