Question

The angle of elevation of the top of a flag post from a point on a horizontal ground is found to be ${30}^o$. On walking $6m$ towards the post, the elevation increased by ${15}^o$. Find the height of the flag post.

Answer 1

Let $y$ be the height of the flag post and $x$ be the initial distance from flag post.

Now, after travelling $6$ m towards the post the distance from the base of pole is $(x-6)$ m.

We know that $\tan \theta =\frac{Oppositeside}{Adjacentside}=\frac{AB}{BC}$

In $△ABC,$ $\angle B={90}^0$ and $\angle C={30}^0$.

Here, $\theta ={30}^0$, $BC=x$ m and $AB=y$ m, therefore,

$\tan \theta =\frac{AB}{BC}\Rightarrow \tan {30}^0=\frac{y}{x}\Rightarrow \frac{1}{\sqrt{3}}=\frac{y}{x}(\because \tan {30}^0=\frac{1}{\sqrt{3}})\Rightarrow x=\sqrt{3}y...........\left(1\right)$

Also, in $△ABD,$ $\angle B={90}^0$ and $\angle D={45}^0$.

Here, $\theta ={45}^0$, $BD=(x-6)$ m and $AB=y$ m, therefore,

$\tan \theta =\frac{AB}{BD}\Rightarrow \tan {45}^0=\frac{y}{x-6}\Rightarrow 1=\frac{y}{x-6}(\because \tan {45}^0=1)\Rightarrow x-6=y\Rightarrow x=y+6...........\left(2\right)$

Comparing equations 1 and 2, we get

$\sqrt{3}y=y+6\Rightarrow \sqrt{3}y-y=6\Rightarrow y(\sqrt{3}-1)=6\Rightarrow y=\frac{6}{\sqrt{3}-1}$

Hence, the height of the flag post is $\frac{6}{\sqrt{3}-1}$ m.