Question

In a right angled triangle the hypotenuse is 2 $\sqrt { 2 }$ times the length of the perpendicular drawn from the opposite vertex on the hypotenuse. Then the other two angles are

• A. $\left( \frac { \pi } { 3 } , \frac { \pi } { 6 } \right)$
• B. $\left( \frac { \pi } { 4 } , \frac { \pi } { 4 } \right)$
• C. $\left( \frac { \pi } { 8 } , \frac { 3 \pi } { 8 } \right)$
• D. $\left( \frac { \pi } { 12 } , \frac { 5 \pi } { 12 } \right)$

Answer 1

Answer: (C)

$\left( \frac { \pi } { 8 } , \frac { 3 \pi } { 8 } \right)$

Answer 2

We have CD = p and AB = 2 $\sqrt { 2 }$ p.

Clearly p = a cosθ = b

sinθ

Now, a² + b² =

$( 2 \sqrt { 2 } p ) ^ { 2 }$

⇒

$p ^ { 2 } \left( \frac { 1 } { \sin ^ { 2 } \theta } + \frac { 1 } { \cos ^ { 2 } \theta } \right) = 8 p ^ { 2 }$

⇒ sin 2θ = ± $\frac { 1 } { \sqrt { 2 } }$

⇒ sin2θ =$\frac { 1 } { \sqrt { 2 } }$

(since 0<θ< 90°)

⇒ 2θ = $\frac { \pi } { 4 }$ ⇒ θ = $\frac { \pi } { 8 }$

⇒ the other angle is

$\frac { \pi } { 2 }$ - θ =$\frac { \pi } { 2 }$ - $\frac { \pi } { 8 }$ = $\frac { 3 \pi } { 8 }$

.