Question

If in a right angled triangle the hypotenuse is four times as

long as the perpendicular drawn to it from opposite vertex,

then one of its acute angle is

• A. $15 ^ { \circ }$
• B. $30 ^ { \circ }$
• C. $45 ^ { \circ }$
• D. None of these

Answer 1

Answer: (A)

$15 ^ { \circ }$

Answer 2

If x is length of perpendicular drawn to it from opposite vertex of a right angled triangle, so, length of diagonal $A B = y _ { 1 } + y _ { 2 }$ ........(i)

From $\Delta O C B , y _ { 2 } = x \cot \theta$ and from $\triangle O C A , y _ { 1 } = x \tan \theta$

Put the values in equation (i), then $A B = x ( \tan \theta + \cot \theta )$ .......(ii)

Since, length of hypotenuse = 4 (Length of perpendicular)

$\therefore$ $x ( \tan \theta + \cot \theta ) = 4 x$ ⇒ $\frac { \sin ^ { 2 } \theta + \cos ^ { 2 } \theta } { \sin \theta \cdot \cos \theta } = 4$

⇒ $\sin 2 \theta = \frac { 1 } { 2 }$ ⇒ $2 \theta = 30 ^ { \circ }$ ⇒ $\theta = 15 ^ { \circ }$.

Trick: $\frac { \text { Length of hypotenuse } } { \text { Length of perpend icular drawn from opposite vertex to hypotenuse } } = \frac { 2 } { \sin 2 \theta }$

⇒ $4 = \frac { 2 } { \sin 2 \theta }$ ⇒ $\sin 2 \theta = \frac { 1 } { 2 }$⇒ $\sin 2 \theta$=$\sin 30 ^ { \circ }$ ⇒ $\theta = 15 ^ { \circ }$.