A) \[\cos A:\cos B:\cos \,C\]
B) \[\cos \frac{A}{2}:\cos \frac{B}{2}:\cos \,\frac{C}{2}\]
C) \[\sin \frac{A}{2}:\sin \frac{B}{2}:\sin \,\frac{C}{2}\]
D) \[\sin A:\sin B:\sin C\]
Correct Answer: B
Solution :
If three forces acting on a particle keep it in equilibrium, each is proportional to the sine of the angle between the other two, then This theorem is known as Lami?s theorem. In \[\Delta ABC,\,\,l\] is the in centre. \[\angle BlC=\pi -\left( \frac{B}{2}+\frac{C}{2} \right)\] \[=\pi -\left( \frac{\pi }{2}-\frac{A}{2} \right)=\frac{\pi }{2}\,+\frac{A}{2}\] Similarly \[\angle AlC\,=\frac{\pi }{2}\,+\frac{B}{2}\] and \[\angle AlB\,=\frac{\pi }{2}+\frac{C}{2}\] By Lami's theorem, \[\frac{P}{\sin \,BlC}=\frac{Q}{\sin AlC}\,=\frac{R}{\sin AlB}\]You need to login to perform this action.
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