Question

The angle of elevation of the top of a pillar from a point on the ground is ${15}^0$ on walking 100 m towards the tower, the angle of elevation is found to be ${30}^0$. Calculate the height of the tower (where tan ${15}^0$ = 2 - $\sqrt{3}$.

Answer 1

Let the height of the piller is $xm$ let the distance between the piller and the initial point on the ground is $ym$

Now, In $△ABC$

$\tan {15}^o=\frac{x}{y}$ $(\because \tan \theta =\frac{oppositeside}{Adjacentside})$

$\because x=y\tan {15}^o$

$x=y(2-\sqrt{3}).....\left(1\right)(\tan {15}^o=2-\sqrt{3}given)$

Again, In $△DBC$

$\tan {30}^o=\frac{x}{y-100}$

$x=(y-100)\left(\tan {30}^o\right)$

$x=(y-100)\frac{1}{\sqrt{3}}.....\left(2\right)$

From eqn $\left(1\right)$ & $\left(2\right)$

As $LHS$ are equal then $RHS$ also equal

$\therefore y(2-\sqrt{3})=(y-100)=\frac{1}{\sqrt{3}}$

multiplying by $\sqrt{3}$ both side

$y(2\sqrt{3}-3)=y-100$

$y(2\sqrt{3}-4)=-100$

$y=-\frac{100}{2\sqrt{3}-4}=\frac{100}{4-2\sqrt{3}}$

Now, from eqn $\left(1\right)$ substituting value of $y$ we get

$x=\frac{100}{4-2\sqrt{3}}(2-\sqrt{3})=\frac{100}{2(2-\sqrt{3})}(2-\sqrt{3})=50m$

$x=50m$