Question

If $D$ is the mid-point of $BC$ of a right angled triangle $ABC$, $(\angle BAC={90}^o)$ such that triangle $ADC$ is an equilateral $\Delta$ then $a^2:b^2:c^2$ is?

• A. $1:3:4$
• B. $3:1:4$
• C. $4:3:1$
• D. $4:1:3$

Answer 1

The correct option is D $4:1:3$

Given that $A=\frac{\pi }{2}$
Since, $△ADC$ is an equilateral triangle.
Therefore, $C=\angle DAC=\frac{\pi }{3}$
$B=\pi -A-C=\frac{\pi }{6}$
Now from sine rule: $a^2:b^2:c^2={\sin }^{2}A:{\sin }^{2}B:{\sin }^{2}C={\sin }^2\frac{\pi }{2}:{\sin }^2\frac{\pi }{6}:{\sin }^2\frac{\pi }{3}$
$\Rightarrow a^2:b^2:c^2=1:\frac{1}{4}:\frac{3}{4}=4:1:3$