Question

Sides of a triangle are $\sqrt{2},\sqrt{6},\sqrt{8}$ then find angles?

Answer 1

Let $a,b,c$ be the sides of triangle $ABC$

Let $a:b:c=\sqrt{2}:\sqrt{6}:\sqrt{8}$

$=\sqrt{2}:\sqrt{2}\sqrt{3}:2\sqrt{2}$

$=1:\sqrt{3}:2$

$\therefore a=1,b=\sqrt{3},c=2$

We know that sides are proportional to its angles.

Using cosine rule, we have

$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{3+4-1}{2\times \sqrt{3}\times 2}=\frac{6}{4\sqrt{3}}=\frac{3}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{2}$

$\cos A=\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$

$\therefore \angle A=\frac{\pi }{6}$

$\cos B=\frac{c^2+a^2-b^2}{2ca}=\frac{4+1-3}{2\times 2\times 1}=\frac{2}{4}=\frac{1}{2}$

$\Rightarrow \cos B=\cos \frac{\pi }{3}=\frac{1}{2}$

$\therefore \angle B=\frac{\pi }{3}$

By angle sum property, $A+B+C=\pi$

$\Rightarrow C=\pi -A-B=\pi -\frac{\pi }{6}-\frac{\pi }{3}=\frac{6\pi -\pi -2\pi }{6}=\frac{3\pi }{6}=\frac{\pi }{2}$

$\therefore \angle C=\frac{\pi }{2}$