Question

In a right-angled triangle, if one of the acute angles is double than the other, then prove that the hypotenuse of the triangle is double the smallest side.

Answer 1

Hint: Assume that the smallest angle is x. Hence the other angle of the triangle will be 2x. Use the fact that the sum of angles of a triangle is $180{}^\circ $. Hence form an equation in x. Solve the equation for x and hence find the measure of the smallest angle. Use the fact that the sine of an angle is equal to the ratio of the opposite side to the hypotenuse. Hence find the value of $\sin x$. Use the fact that $\sin 30{}^\circ =\dfrac{1}{2}$ and hence prove that hypotenuse is double the smallest side.

Complete step-by-step solution -

Let $\angle C=x$
Hence, we have $\angle B=2x$
Now, we know by angle sum property of a triangle
$\begin{align}
  & x+2x+90{}^\circ =180{}^\circ \\
 & \Rightarrow 3x+90{}^\circ =180{}^\circ \\
\end{align}$
Subtracting 90 from both sides of the equation, we get
$3x=90{}^\circ $
Dividing by 3 on both sides of the equation, we get
$x=30{}^\circ $
Hence, we have x = 30
Now, in triangle ABC, AB is the side opposite to C and BC is the hypotenuse. Hence, we have
$\sin x=\dfrac{AB}{BC}$
Substituting the value of x, we get
$\sin 30{}^\circ =\dfrac{AB}{BC}$
We know that $\sin 30{}^\circ =\dfrac{1}{2}$
Hence, we have
$\dfrac{AB}{BC}=\dfrac{1}{2}$
Cross multiplying, we get
BC = 2AB.
Also, since C is the smallest angle, AB is the smallest side(The side opposite to a larger angle is longer)
Hence the hypotenuse is double the smallest side.

Note: Alternative Solution:
By sine rule, we have $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$
Hence, we have
$\dfrac{BC}{\sin \left( 90{}^\circ \right)}=\dfrac{AB}{\sin \left( 30{}^\circ \right)}$
Hence, we have
$BC=AB$
Hence proved.