Question

Two points A and B are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are 60° and 45° respectively. If the height of the tower is $15$ m, then find the distance between the points.

Answer 1

Let CD be the tower. A and B are the two points on the same side of the tower.



In $\Delta DBC$,

$\tan 60$=$\frac{DC}{BC}$

$\Rightarrow \sqrt{3}$=$\frac{15}{BC}$

$\Rightarrow BC$=$\frac{15}{\sqrt{3}}$

$\Rightarrow BC$=$5\sqrt{3}$ m
In $\Delta DAC$,

$\tan 45°$=$\frac{DC}{AC}$

$\Rightarrow 1$=$\frac{15}{AC}$

$\Rightarrow$ $AC$=$15$ m
Now,
AC = AB + BC
∴ AB = AC − BC

$=15-5\sqrt{3}$

$=5\sqrt{3}(\sqrt{3}-1)$ m

Hence, the distance between the two points A and B is $5(3-\sqrt{3})$ m or $5\sqrt{3}(\sqrt{3}-1)$m.