question_answer

Point D, E are taken on the side BC of a triangle \[ABC\]such that \[BD=DE=EC\].If \[\angle BAD=x\], \[\angle DAE=y\], \[\angle EAC=z\], then the value of \[\frac{\sin (x+y)\sin (y+z)}{\sin x\sin z}=\]

A) 1

B) 2

C) 4

D) None of these

Correct Answer: C

Solution :

From\[\Delta ADC\], \[\frac{\sin (y+z)}{DC}=\frac{\sin C}{AD}\] From \[\Delta ABD,\frac{\sin x}{BD}=\frac{\sin B}{AD}\] From \[\Delta AEC,\frac{\sin z}{EC}=\frac{\sin C}{AE}\] From \[\Delta ABE,\,\frac{\sin (x+y)}{BE}=\frac{\sin B}{AE}\] Therefore\[\frac{\sin (x+y)\sin (y+z)}{\sin x\sin z}\] \[=\frac{BE}{AE}\times \frac{DC}{AD}\times \frac{AD}{BD}\times \frac{AE}{EC}=\frac{2BD\times 2EC}{BD\times EC}=4\].