Question

The sides of a triangle are in the ratio $1:\sqrt{3}:2$. Then the angles are in the ratio

• A. $1:2:3$
• B. $1:2:4$
• C. $1:4:5$
• D. $1:3:5$

Answer 1

The correct option is A $1:2:3$

Let the sides be represented by

$a=k;b=k\sqrt{3};c=2k$

and their corresponding angle of the opposite vertices by represented by $A,B,C$

Using cosine rule,

$cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{3k^2+4k^2-k^2}{4k^2\sqrt{3}}=\frac{\sqrt{3}}{2}$

=> $\angle A=\pi /6$

Similarly, $cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{1}{2}$

=> $\angle B=\pi /3$

$cosC=\frac{a^2+b^2-c^2}{2ab}=0$

=> $\angle C=\pi /2$

=> $A:B:C=1:2:3$