Question

In a right-angled triangle, one acute angle is double the other, then the hypotenuse is

• A. Equal to the smallest side
• B. Double the smallest side
• C. Triple the smallest side
• D. None of these

Answer 1

The correct option is B

Double the smallest side

Step-1: Draw $\mathbf{∆}\mathbf{ABC}$

Let $\angle \mathrm{BAC}=\mathrm{x}$

then $\angle \mathrm{ACB}=2\mathrm{x}$

$\angle \mathrm{ABC}=90°$

Step-2: Find the smallest side of $\mathbf{∆}\mathbf{ABC}$

In $∆\mathrm{ABC}$,

$\angle \mathrm{ABC}+\angle \mathrm{BAC}+\angle \mathrm{ACB}=180°.......\left(1\right)$ ( Angle Sum Property )

On putting values in $\left(1\right)$

$\Rightarrow 90°+\mathrm{x}+2\mathrm{x}=180°$

$\Rightarrow \mathrm{x}=\angle \mathrm{BAC}=30°$

$\Rightarrow \angle \mathrm{ACB}=2\times 30°=60°$

The side opposite the smallest angle is the smallest.

Hence, BC is the smallest side of the triangle.

Step-3: Find the relation between the hypotenuse and the smallest side.

In $∆\mathrm{ABC}$,

$\cos \left(2\mathrm{x}\right)=\frac{\mathrm{BC}}{\mathrm{AC}}$

$\Rightarrow \cos \left(60°\right)=\frac{\mathrm{BC}}{\mathrm{AC}}$

$\Rightarrow \frac{1}{2}=\frac{\mathrm{BC}}{\mathrm{AC}}\left[\therefore \cos \left(60°\right)=\frac{1}{2}\right]$

$\mathrm{AC}=2·\mathrm{BC}$

So, Hypotenuse is double the smallest side.

Hence, option $\left(\mathrm{B}\right)$ is the correct answer.