Question

An observed standing $40m$ away from a building observed that the angle of elevation of the top and bottom of a flagstaff, which is surmounted on the building are ${60}^o$ and ${45}^o$ respectively. Find the height of the building and the length of the flagstaff $(\sqrt{3}=1.732)$

Answer 1

In $\Delta$ BCD

$\frac{BC}{DC}=\tan {45}^0$

$\Rightarrow \frac{BC}{40}=1$

$\Rightarrow BC=40m$

In $\Delta$ACD

$\Rightarrow \frac{AC}{DC}=\tan {60}^0$

$\Rightarrow \frac{AB+BC}{40}=\sqrt{3}fromequation\left(1\right)$

$\Rightarrow AB=40\sqrt{3}-40$

$\Rightarrow AB=40\left(\sqrt{3}-1\right)$

$\Rightarrow AB=40\left(1.732-1\right)$

$\Rightarrow AB=40\left(0.732\right)$

$\Rightarrow AB=40\times 7.32$

$AC=AB+BC$

$=29.28+40$

$=69.28m$
Therefore height of building is $96.28m$ and height of flag staff is $29.28m$