The sides of a triangle are in the ratio $1:\sqrt 3: 2$, then the angles are in the ration-

Question

Using the sine rule

$$\sin A:\sin B:\sin C=1:\sqrt 3: 2$$

This is small question, so I could only do one step. How should finish it?

Answer 1

The best answer is in comment. In any case it is simpler to find the angles themselves and use the result to find the ratio.

In general case use the cosine theorem to find the angles in a triangle: $$ \cos A=\frac{b^2+c^2-a^2}{2bc}. $$ In your case one obtains: $$ \cos A=\frac{\sqrt3}2,\;\cos B=\frac12,\; \cos C=0. $$ I assume the angles with these cosines are well-known to you.

Answer 2

In a triangle, given that $$a=k, b=\sqrt{3} k, c=2k \implies c^2=a^2+b^2, ~\text{so}~ C=\pi/2$$ Next, bt sin-law: $$\frac{\sin A}{1}=\frac{\sin B}{\sqrt{3}}=\frac{\sin C}{2} \implies \frac{\sin A}{1/2}=\frac{\sin B}{\sqrt{3}/2}=\frac{\sin C}{1} \implies A=\pi/6, B=\pi/3, C=\pi/2.$$ Also check that $A+B+C=\pi.$ So $$A:B:C:=1:2:3$$