A) \[12+2\sqrt{3}\]
B) \[12-2\sqrt{3}\]
C) \[2\sqrt{3}+2\]
D) \[2\sqrt{3}-2\]
Correct Answer: A
Solution :
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;\[\sin C=\sin ({{180}^{o}}-(A+B))=\sin (A+B)\] \[=\sin A\cdot \cos B+\cos A\cdot \sin B\] \[=\left( \frac{6\sqrt{3}}{11} \right)\left( \frac{1}{2} \right)+\sqrt{\left\{ 1-{{\left( \frac{6\sqrt{3}}{11} \right)}^{2}} \right\}}\cdot \left( \frac{\sqrt{3}}{2} \right)\] \[=\frac{1}{22}\left( \frac{\sqrt{3}}{2} \right)(12+2\sqrt{3})\] Now,\[c=\frac{b}{\sin B}\sin C\] \[\Rightarrow \,\,c=12+2\sqrt{3}\].
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