From a point $ P $ on the ground the angle of elevation of a $ 10 \mathrm{~m} $ tall building is $ 30^{\circ} $. A flag is hoisted at the top of the building and the angle of elevation of the top of the flag-staff from $ P $ is $ 45^{\circ} $. Find the length of the flag-staff and the distance of the building from the point $ P $. (Take $ \sqrt{3}=1.732 $ ).

Given:

From a point \( P \) on the ground the angle of elevation of a \( 10 \mathrm{~m} \) tall building is \( 30^{\circ} \).

A flag is hoisted at the top of the building and the angle of elevation of the top of the flag-staff from \( P \) is \( 45^{\circ} \).

To do:

We have to find the length of the flag-staff and the distance of the building from the point \( P \).

Solution:  

Let $AB$ be the tall building and $BC$ be the length of the flag-staff.

Point $P$ be the point of observation.

From the figure,

$\mathrm{AB}=10 \mathrm{~m}, \angle \mathrm{CPA}=45^{\circ}, \angle \mathrm{BPA}=30^{\circ}$

Let the height of the flag staff be $\mathrm{BC}=h \mathrm{~m}$ and the distance of the building from the point $P$ be $\mathrm{AP}=x \mathrm{~m}$.

This implies,

$\mathrm{AC}=10+h \mathrm{~m}$

We know that,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { CA }}{PA}$

$\Rightarrow \tan 45^{\circ}=\frac{10+h}{x}$

$\Rightarrow 1=\frac{10+h}{x}$

$\Rightarrow x(1)=10+h \mathrm{~m}$

$\Rightarrow x=10+h \mathrm{~m}$.........(i)

Similarly,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { BA }}{PA}$

$\Rightarrow \tan 30^{\circ}=\frac{10}{x}$

$\Rightarrow \frac{1}{\sqrt3}=\frac{10}{x}$

$\Rightarrow x=10\sqrt3=10(1.732)=17.32 \mathrm{~m}$

$\Rightarrow 10+h=10\sqrt3 \mathrm{~m}$            [From (i)]

$\Rightarrow h=10(1.732-1) \mathrm{~m}$

$\Rightarrow h=10(0.732) \mathrm{~m}$

$\Rightarrow h=7.32 \mathrm{~m}$

Therefore, the height of the flag-staff is $7.32 \mathrm{~m}$ and the distance of the building from point $P$ is $17.32 \mathrm{~m}$.    

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Updated on: 10-Oct-2022

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