Question

If the hypotenuse of a right - angled triangle is four times the length of the perpendicular drawn from the opposite vertex to it, then the difference of the two acute angles will be

• A. ${60}^0$
• B. ${15}^0$
• C. ${75}^0$
• D. ${30}^0$

Answer 1

The correct option is A ${60}^0$

Consider the given figure.
$\sin C=\frac{DB}{BC}$
$=\frac{x}{4x\cos C}$
Hence, $\sin C.\cos C=\frac{1}{4}$
$2\sin C.\cos C=\frac{1}{2}$
$\sin 2C=\sin {30}^0$
$C={15}^0$
Hence
$A={90}^0-{15}^0={75}^0$
Hence, $A-C$
$={60}^0$