Question

lf the sides of a triangle are $\sqrt{2},\sqrt{6},\sqrt{8}$ then the angles of the triangle are

• A. ${30}^0,{75}^0,{75}^0$
• B. ${45}^0,{45}^0,{90}^0$
• C. ${30}^0,{60}^0,{90}^0$
• D. ${60}^0,{60}^0,{60}^0$

Answer 1

The correct option is C ${30}^0,{60}^0,{90}^0$

Given, $a=\sqrt{2},b=\sqrt{6},c=\sqrt{8}$

$\therefore a^2+b^2$$=(\sqrt{2}{)}^2+(\sqrt{6}{)}^2$

$=2+6$ $=8$ $=(\sqrt{8}{)}^2$ $=c^2$

$\to c^2=a^2+b^2$

Hence the above triangle is a right angled triangle.

Let, $a=\sqrt{2}$

$b=\sqrt{6}$

$c=\sqrt{8}$

Then, $\angle C={90}^0$

And $\sin A=\frac{\sqrt{2}}{\sqrt{8}}$ $=\frac{1}{2}$

Hence, $\angle A={30}^0$

Therefore $\angle B$ $={180}^0-({90}^0+{30}^0)$ $={60}^0$.