NCERT Class X Chapter 9: Some Application Of Trigonometry Example 4
NCERT Class X Chapter 9: Some Applications of Trigonometry Example 4
Question:
From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (You may take \( \sqrt{3} = 1.732 \))
Given:
- Height of the building = 10 m
- Angle of elevation of the top of the building from P = \( 30^\circ \)
- Angle of elevation of the top of the flagstaff from P = \( 45^\circ \)
- \( \sqrt{3} = 1.732 \) (to be used in calculation)
To Find:
- Length of the flagstaff
- Distance of the building from point P
Formula:
Trigonometric ratio for tangent:
$$ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} $$Solution:
Step 1: Let the distance from point P to the base of the building be \( x \) meters. Let the length of the flagstaff be \( h \) meters.
Step 2: For the top of the building (without flag), the angle of elevation is \( 30^\circ \).
$$ \tan 30^\circ = \frac{10}{x} $$Step 3: We know \( \tan 30^\circ = \frac{1}{\sqrt{3}} \). Substitute and solve for \( x \):
$$ \frac{1}{\sqrt{3}} = \frac{10}{x} $$ $$ x = 10 \sqrt{3} $$Step 4: For the top of the flagstaff, the angle of elevation is \( 45^\circ \). The total height is \( 10 + h \) meters.
$$ \tan 45^\circ = \frac{10 + h}{x} $$Step 5: We know \( \tan 45^\circ = 1 \). Substitute and solve for \( h \):
$$ 1 = \frac{10 + h}{x} $$ $$ x = 10 + h $$Step 6: Substitute the value of \( x \) from Step 3:
$$ 10\sqrt{3} = 10 + h $$ $$ h = 10\sqrt{3} - 10 $$Step 7: Substitute \( \sqrt{3} = 1.732 \) and calculate the values:
$$ x = 10 \times 1.732 = 17.32 \text{ m} $$ $$ h = 10 \times (1.732 - 1) = 10 \times 0.732 = 7.32 \text{ m} $$Result:
Length of the flagstaff \( \approx 7.32 \) m
Distance of the building from point P \( = 10\sqrt{3} \) m \( \approx 17.32 \) m
Comments
Post a Comment