Question

The angle of elevation of the top of a tower from a point A on the ground is $30°$ on moving a distance of $20\text{m}$ towards the foot of the tower to a point B the angle of elevation increases to $60°$ find the height of the tower and the distance of the tower from the point A.

Answer 1

Step 1: Draw the figure for the given question :

Step 2: Calculate the height of the tower w.r.t $x$ :

In triangle BCD,

$\begin{array}{c}\tan 60°=\frac{CD}{BC}\\ \sqrt{3}=\frac{h}{x}\\ h=x\sqrt{3}\mathrm{.........}\left(1\right)\end{array}$

Step 3: Calculate the distance of the tower from the point A :

In triangle ACD,

$\begin{array}{c}\tan 30°=\frac{h}{x+20}\\ \frac{1}{\sqrt{3}}=\frac{x\sqrt{3}}{x+20}\\ x\times 3=x+20\\ 3x=x+20\\ 3x-x=20\\ 2x=20\\ x=\frac{20}{2}\\ x=10\end{array}$

Therefore, the distance of the tower from the point A:

$20+10=30\text{\hspace{0.17em}m}$

Step 4: Calculate the height of the tower :

Therefore, the height of the tower:

$\begin{array}{c}h=x\sqrt{3}\\ h=10\times \sqrt{3}\\ h=10\times 1.732\\ h=17.32\text{\hspace{0.17em}m}\end{array}$

Hence, the distance of the tower from the point A is $30\text{m}$ and the height of the tower is $17.32\text{m}$.