Question

In a triangle, the length of the two larger sides are $24$ and $22$, respectively. If the angles are in AP, then the third side is

• A. $12+2\sqrt{3}$
• B. $12-2\sqrt{3}$
• C. $2\sqrt{3}+2$
• D. $2\sqrt{3}-2$

Answer 1

The correct option is A $12+2\sqrt{3}$

Since the angles are in AP, therefore
$2B=A+C\Rightarrow 3B=A+B+C$
$\Rightarrow 3B={180}^o$ $\Rightarrow B={60}^o$
using the formula,
$\frac{\sin A}{a}=\frac{\sin B}{b}$, we get
$\sin A=\frac{a}{b}\sin B=\frac{24}{22}\cdot \sin {60}^o=\frac{6\sqrt{3}}{11}$
Now, $\sin C=\sin ({180}^o-(A+B))=\sin (A+B)$
$=\sin A\cdot \cos B+\cos A\cdot \sin B$
$=\left(\frac{6\sqrt{3}}{11}\right)\left(\frac{1}{2}\right)+\sqrt{\{1-{\left(\frac{6\sqrt{3}}{11}\right)}^2\}}\cdot \left(\frac{\sqrt{3}}{2}\right)$
$=\frac{1}{22}\left(\frac{\sqrt{3}}{2}\right)(12+2\sqrt{3})$
Now, $c=\frac{b}{\sin B}\sin C$
$\Rightarrow c=12+2\sqrt{3}$.