Question

From a point P on the ground, the angle of elevation of a 10m tall building is ${{30}^{o}}$. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is ${{45}^{o}}$. Find the length of the flagstaff and the distance of the building from point P.

Answer 1

Hint: First of all, draw a building AB of 10m and its angle of elevation at point P as ${{30}^{o}}$. Now draw a flagstaff CA and angle of elevation of its top at point P as ${{45}^{o}}$. Now take $\tan {{30}^{o}}\text{ and }\tan {{45}^{o}}$ and use $\tan \theta =\dfrac{perpendicular}{base}$ on these two angles to get the required values.

Complete step-by-step answer:
In this question, we are given that from a point P on the ground the angle of elevation of a 10m tall building is ${{30}^{o}}$. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is ${{45}^{o}}$. Here, we have to find the length of the flagstaff and the distance of the building from point P.
First of all, let us draw a 10m long building AB whose angle of elevation from point P on the ground is ${{30}^{o}}$.
Now, let us draw a flagstaff CA on top of the building whose angle of elevation at point P is ${{45}^{o}}$.

Let us consider the length of the flagstaff CA = h and distance of the building from point P, i.e. BP = d. We know that,
$\tan \theta =\dfrac{perpendicular}{base}$
In triangle ABP,
$\tan \left( \angle APB \right)=\dfrac{AB}{BP}....\left( i \right)$
We are given that,
$\angle APB={{30}^{o}}$
$AB=10m$
$BP=d$
By substituting these values in equation (i), we get,
$\tan {{30}^{o}}=\dfrac{10}{d}....\left( ii \right)$
In triangle CBP,
$\tan \left( \angle CPB \right)=\dfrac{CB}{BP}....\left( iii \right)$
We are given that,
$\angle CPB={{45}^{o}}$
$CB=CA+AB=h+10m$
$BP=d$
By substituting these values in equation (iii), we get,
$\tan {{45}^{o}}=\dfrac{h+10}{d}....\left( iv \right)$
By dividing equation (ii) by equation (iv), we get,
$\dfrac{\tan {{30}^{o}}}{\tan {{45}^{o}}}=\dfrac{\dfrac{10}{d}}{\dfrac{h+10}{d}}$
$\dfrac{\tan {{30}^{o}}}{\tan {{45}^{o}}}=\left( \dfrac{10}{d} \right)\left( \dfrac{d}{h+10} \right)$
By canceling the like terms, we get,
$\dfrac{\tan {{30}^{o}}}{\tan {{45}^{o}}}=\dfrac{10}{h+10}....\left( v \right)$
Now, let us find the value of $\tan {{30}^{o}}\text{ and }\tan {{45}^{o}}$ from the trigonometric table of general angles.

| $\sin \theta$| $\cos \theta$| $\tan \theta$| $\operatorname{cosec}\theta$| $\sec \theta$| $\cot \theta$
---|---|---|---|---|---|---
0| 0| 1| 0| -| 1| -
$\dfrac{\pi }{6}$| $\dfrac{1}{2}$| $\dfrac{\sqrt{3}}{2}$| $\dfrac{1}{\sqrt{3}}$| 2| $\dfrac{2}{\sqrt{3}}$| $\sqrt{3}$
$\dfrac{\pi }{4}$| $\dfrac{1}{\sqrt{2}}$| $\dfrac{1}{\sqrt{2}}$| 1| $\sqrt{2}$| $\sqrt{2}$| 1
$\dfrac{\pi }{3}$| $\dfrac{\sqrt{3}}{2}$| $\dfrac{1}{2}$| $\sqrt{3}$| $\dfrac{2}{\sqrt{3}}$| 2| $\dfrac{1}{\sqrt{3}}$
$\dfrac{\pi }{2}$| 1| 0| -| 1| -| 0

From the above table, we get,
$\tan {{30}^{o}}=\dfrac{1}{\sqrt{3}}$
$\tan {{45}^{o}}=1$
By substituting these values in equation (v), we get,
$\dfrac{\dfrac{1}{\sqrt{3}}}{1}=\dfrac{10}{h+10}$
$\dfrac{1}{\sqrt{3}}=\dfrac{10}{h+10}$
By cross multiplying the above equation, we get,
$h+10=10\sqrt{3}$
$h=10\sqrt{3}-10$
$h=10\left( \sqrt{3}-1 \right)m$
We know that,
$\sqrt{3}\simeq 1.732$
So, we get,
$h=10\left( 1.732-1 \right)$
$h=10\left( 0.732 \right)$
$h=7.32m$
Now considering equation (ii),
$\tan {{30}^{o}}=\dfrac{10}{d}$
We know that, $\tan {{30}^{o}}=\dfrac{1}{\sqrt{3}}$. So, we get,
$\dfrac{1}{\sqrt{3}}=\dfrac{10}{d}$
$d=10\sqrt{3}m$
$d=10\left( 1.732 \right)$
$d\approx 17.32m$
Hence, we get the length of the flagstaff as 7.32m and distance between the building and point P is 17.32m.
Note: In this question, students often make the mistake of taking $\angle APC={{45}^{o}}$ which is wrong because it is written that from the top of the flagstaff, the angle of elevation is ${{45}^{o}}$ at point P and not that flagstaff is subtending angle ${{45}^{o}}$ at point P. So, this must be taken care of and students should properly read the question and draw the diagram first to visualize these types of questions.