Question

From the top of a 7 m height building, the angle pf elevation of the top of a tower is ${60}^{\circ }$ and angle of depression of the tower is ${45}^{\circ }$. Find the height of the tower.

Answer 1

Let AB be the building and tower be CE.

Given height of building AB$=7$m

From the top o building, angle of elevation of top of tower$={60}^o$

$\angle EAG={60}^o$

Angle of depression of the foot of the tower$={45}^o$

$\angle CAD={45}^o$

$CE=?$

Since AB & CD are parallel,

$CD=AB=7m$

Also AD & BC are parallel

$AD=BC$

Since tower & building are vertical to ground

$\angle ABC={90}^o$ & $\angle EDA={90}^o$

AD & BC are parallel, AC is traversal

$\angle ACB=\angle DAC$ (alternate angles)

$\angle ACB={45}^o$

In right $\Delta ABC$

$\tan c=$ side opp to angle c/side adjacent to angle c.

$\tan {45}^o=AB/BC$

$1=AB/BC$

$1=7/BC$

$BC=7$m

Since $BC=AD$, $AD=7$cm

In right $\Delta ADE$

$\tan 60=ED/AD$

$\sqrt{3}=ED/AD$

$\sqrt{3}=ED/7$

$ED=7\sqrt{3}$

Height of tower $=ED+DC$

$=7\sqrt{3}+7$

$7(\sqrt{3}+1)m$.