Question

lf the angles of a triangle are in $A.P$. with common difference equal to $\frac{1}{3}$ of the least angle,

then the sides are in the ratio is?

• A. $\sqrt{2}:2\sqrt{3}:\sqrt{6}+\sqrt{2}$
• B. $2\sqrt{2}:\sqrt{3}:\sqrt{6}-\sqrt{2}$
• C. $2\sqrt{2}:2\sqrt{3}:\sqrt{6}-\sqrt{2}$
• D. $2\sqrt{2}:2\sqrt{3}:\sqrt{6}+\sqrt{2}$

Answer 1

The correct option is D $2\sqrt{2}:2\sqrt{3}:\sqrt{6}+\sqrt{2}$

Let us consider the least angle be angle $A$

So, Common difference be $\frac{A}{3}$

hence ,

$B=\frac{4A}{3}$

$C=\frac{7A}{3}$

sum of the angles are equal to ${180}^{\circ }$

$A+\frac{4A}{3}+\frac{7A}{3}={180}^{\circ }$

$A=\frac{2\pi }{7}$, $B=\frac{8\pi }{21}$, and $C=\frac{2\pi }{3}$

using the sine rule

$a:b:c=\sin A:\sin B:\sin C$

$a:b:c=2\sqrt{2}:2\sqrt{3}:\sqrt{6}+\sqrt{2}$