Question

A flag-staff a metre high placed on the top of a tower b metre high subtends the same angle as the tower to an observer h metre high standing on a horizontal plane at a distance d metres from the foot of the tower. Prove that
$(a-b)d^2=(a+b)b^2-2b^{2}h-(a-b)h^2$

Answer 1

BM is bisector of $\angle AMC$ and hence it divides the opposite side in the ration of the arms of the angle.
$\frac{b}{a}=\frac{AM}{CM}$ or $\frac{b^2}{a^2}=\frac{d^2+h^2}{d^2+(a+b-h{)}^2}$
$\therefore a^2(d^2+h^2)=b^2[d^2+(a+b{)}^2+h^2-2h(a+b)]$
or $\therefore d^2(a^2-b^2)=-h^2(a^2-b^2)+b^2(a+b{)}^2-2h{b}^2(a+b)$
Cancel a+b etc and you get the result.