NIMCET 2015 — Mathematics PYQ

Let the right-angled triangle be $△ABC$ with $\angle B=90^{\circ }$. Let the hypotenuse $AC=h$ and the perpendicular from $B$ to $AC$ be $BD=p$.

Given: $h=4p⟹p=\frac{h}{4}$

In $△ABC$:

$AB=h\cos A$

$BC=h\sin A$

Area of $△ABC=\frac{1}{2}\times AB\times BC$

$\text{Area}=\frac{1}{2}\times (h\cos A)\times (h\sin A)=\frac{1}{2}{h}^2\sin A\cos A$ — (1)

Also, Area of $△ABC=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times AC\times BD$

$\text{Area}=\frac{1}{2}\times h\times p$

$\text{Area}=\frac{1}{2}\times h\times \frac{h}{4}=\frac{h^2}{8}$ — (2)

Equating (1) and (2):

$\frac{1}{2}{h}^2\sin A\cos A=\frac{h^2}{8}$

$\sin A\cos A=\frac{1}{4}$

Multiply both sides by $2$:

$2\sin A\cos A=\frac{2}{4}$

$\sin 2A=\frac{1}{2}$

Since $\sin 30^{\circ }=\frac{1}{2}$:

$2A=30^{\circ }$ or $2A=150^{\circ }$

$A=15^{\circ }$ or $A=75^{\circ }$

Correct Option: (c)