Find the acute angles of a right angled triangle whose hypotenuse is four times as long as the perpendicular drawn to it from the opposite angle.
Let $ABC$ be a right angled triangle with right angle at $B$ and sides $a, \; b, \; c$ as shown in the figure.
$BP$ is the perpendicular from $B$ to the side $b$.
Given: $\;$ $b = 4 \; BP$ $\implies$ $BP = \dfrac{b}{4}$
From the figure,
in $\triangle ABC$, $\;$ $\cos C = \dfrac{a}{b}$ $\;\;\; \cdots \; (1)$
in $\triangle BPC$, $\;$ $\sin C = \dfrac{b / 4}{a} = \dfrac{b}{4a}$ $\;\;\; \cdots \; (2)$
From equations $(1)$ and $(2)$,
$\sin C \times \cos C = \dfrac{b}{4a} \times \dfrac{a}{b}$
i.e. $\;$ $\sin C \cos C = \dfrac{1}{4}$
$\therefore \;$ $2 \sin C \cos C = 2 \times \dfrac{1}{4} = \dfrac{1}{2}$
i.e. $\;$ $\sin 2 C = \dfrac{1}{2}$
i.e. $\;$ $2 C = 30^{\circ}$ $\;$ or $\;$ $2 C = 150^{\circ}$
i.e. $C = 15^{\circ}$ $\;$ or $\;$ $C = 75^{\circ}$
When $\;$ $C = 15^{\circ}$, $\;$ $A = 180^{\circ} - \left(B + C\right) = 180^{\circ} - \left(90^{\circ} + 15^{\circ}\right) = 75^{\circ}$
When $\;$ $C = 75^{\circ}$, $\;$ $A = 180^{\circ} - \left(B + C\right) = 180^{\circ} - \left(90^{\circ} + 75^{\circ}\right) = 15^{\circ}$
$\therefore \;$ The angles of the triangle are
$A = 75^{\circ}$, $\;$ $B = 90^{\circ}$, $\;$ $C = 15^{\circ}$ $\;$ or $\;$ $A = 15^{\circ}$, $\;$ $B = 90^{\circ}$, $\;$ $C = 75^{\circ}$