Question

A tower stands vertically in a field. The field is in the shape of an equilateral triangle of side 100 metres. The tower subtends angles of ${45}^o,{60}^o,$ and ${60}^o$ at the vertices of the triangle. Find the height of tower.

Answer 1

Let the tower stand at O and its height OP = h which subtends an angle of ${45}^o,{60}^o,{60}^o$ at A, B, C respectively.
$OA=hcot{45}^o=h$;
$OB=hcot{60}^o=\frac{h}{\sqrt{3}}=OC$
$OB=OC$
Also AB = AC = BC = a say.
If D be the mid-point of BC then OD and AD both are perpendicular to base BC.
$\therefore$ AD is median as well as altitude. In an isosceles or equilateral $\Delta$, both centroid and orthocentre coincide, then
$OA=\frac{2}{3}AD=\frac{2}{3}\sqrt{a^2+\frac{a^2}{4}}$
or $h=OA=\frac{2}{3}.\frac{1}{2}\sqrt{5}a$
$=\frac{100\sqrt{5}}{3}\therefore a=100$, given.