Question

A statue, $1.6m$ tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of statue is ${60}^{\circ }$ and from the same point the angle of elevation of the top of the pedestal is ${45}^{\circ }$. Find the height of the pedestal.

Answer 1

Let $AB$ be the statue, $BC$ be the pedestal, and $D$ be the point on the ground from where the elevation angles are to be measured.

In $\Delta BCD$, we have

$\tan {45}^{\circ }=\frac{BC}{CD}$ [ $\because \tan \theta =\frac{\text{opposite side}}{\text{Adjacent side}}$]

$\Rightarrow \frac{BC}{CD}=1$ [$\because \tan {45}^{\circ }=1$]

$\therefore BC=CD......\left(i\right)$

In $\Delta ACD$, we have
$\tan {60}^{\circ }=\frac{AB+BC}{CD}$

$\sqrt{3}=\frac{AB+BC}{CD}$ [$\because \tan {60}^{\circ }=\sqrt{3}$]

$\sqrt{3}CD=AB+BC$

$\sqrt{3}BC=1.6+BC$ [Since, $CD=BC$ and $AB=1.6$]

$BC\left(\sqrt{3}\right)-BC=1.6$

$BC(\sqrt{3}-1)=1.6$

$BC=\frac{1.6}{(\sqrt{3}-1)}$

Rationalizing the denominator:
$BC=\frac{1.6}{(\sqrt{3}-1)}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$=\frac{1.6(\sqrt{3}+1)}{(\sqrt{3}{)}^2-1^2}$ [$\because (a^2-b^2)=(a-b)(a+b)$]

$=\frac{1.6(\sqrt{3}+1)}{3-1}$

$=\frac{1.6(\sqrt{3}+1)}{2}$

$\therefore BC=0.8(\sqrt{3}+1)m$

Therefore, the height of the pedestal is $0.8(\sqrt{3}+1)m$.