Question

In a right angled triangle the hypotenuse is four times the length of the perpendicular drawn from the opposite vertex on the hypotenuse, then one of the other angle is

• A. ${15}^{\circ }$
• B. ${75}^{\circ }$
• C. $22{\frac{1}{2}}^{\circ }$
• D. $67{\frac{1}{2}}^{\circ }$

Answer 1

Answer: (A), (B)

The correct options are
A ${15}^{\circ }$
B ${75}^{\circ }$

In $△ABC$ Let $\angle ACB=\theta ,BD=x\Rightarrow AC=4x$

And $\cos \theta =\frac{BC}{AC}=\frac{BC}{4x}.......\left(1\right)$

In $△BCD,\sin \theta =\frac{BD}{BC}=\frac{x}{BC}.....\left(2\right)$ From $\left(1\right)$ and $\left(2\right)$, we get,

$\sin \theta \cos \theta =\frac{1}{4}\Rightarrow \sin 2\theta ={30}^{\circ }\Rightarrow \theta ={15}^{\circ }$

Therefore, $\angle BAC={90}^{\circ }-{15}^{\circ }={75}^{\circ }$