Q131 / 200 · Manipal (MET) 2013 · Mathematics
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
- Correct answerA$12+2 \sqrt{3}$
- B$12-2 \sqrt{3}$
- C$2 \sqrt{3}+2$
- D$2 \sqrt{3}-2$
Solution
Since the angles are in $A P$, therefore
$2 B=A+C \quad \Rightarrow \quad 3 B=A+B+C$
$\Rightarrow 3 B=180^{\circ} \Rightarrow B=60^{\circ}$
using the formula,
$\frac{\sin A}{a}=\frac{\sin B}{b}$, we get
$\sin A=\frac{a}{b} \sin B=\frac{24}{22} \cdot \sin 60^{\circ}=\frac{6 \sqrt{3}}{11}$
Now; $\quad \sin C=\sin \left(180^{\circ}-(A+B)\right)=\sin (A+B)$
$=\left(\frac{6 \sqrt{3}}{11}\right)\left(\frac{1}{2}\right)+\sqrt{\left\{1-\left(\frac{6 \sqrt{3}}{11}\right)^2\right\}} \cdot\left(\frac{\sqrt{3}}{2}\right)$
$=\frac{1}{22}\left(\frac{\sqrt{3}}{2}\right)(12+2 \sqrt{3})$
Now, $c=\frac{b}{\sin B} \sin C$
$\Rightarrow \quad c=12+2 \sqrt{3}$.
Hence, the correct answer is option: 'A'.