Question

The angle of elevation of the top of a tower as observed from a point in a horizontal plane through the foot of the tower is ${32}^o$. When the observer moves towards the tower a distance of 100 m, he finds the angle of elevation of the top to be ${63}^o$. Find the height of the tower and the distance of the first position from the tower. [Take $\tan {32}^{\circ }=0.6248$ and $\tan {63}^{\circ }=1.9626$]

Answer 1

Let the Height of tower be $AB$
in $△ABC$
$\tan {63}^o=\frac{AB}{BC}\Rightarrow \frac{AB}{1.96}=BC...........\left(1\right)$
In $△ABD$
$\tan {32}^o=\frac{AB}{BD}\Rightarrow BD=\frac{AB}{0.62}............\left(2\right)$
and it is given that $CD=100m=BD-BC$

From $\left(1\right)$ and $\left(2\right)$

$\Rightarrow \frac{AB}{0.62}-\frac{AB}{1.96}=100$

$\Rightarrow 1.34AB=121.52$

$\Rightarrow AB=\frac{121.52}{1.34}=90.68m$

And

$BC=\frac{AB}{1.96}=\frac{90.68}{1.96}=46.26m$