question_answer

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

A)  1                            

B)         2                            

C)         4                            

D)         none of these

Correct Answer: C

Solution :

From \[\Delta ADC,\frac{\sin (y+z)}{DC}=\frac{\sin C}{AD}\] (Applying sine rule)                 From \[\Delta \Alpha \Beta D,\]\[\frac{\sin x}{BD}=\frac{\sin B}{AD}\]                 From\[\Delta AEC,\]\[\frac{\sin z}{EC}=\frac{\sin C}{AE}\]                 From \[\Delta ABE,\frac{\sin (x+y)}{BE}=\frac{\sin B}{AE}\]                                 \[\therefore \]  \[\frac{\sin (x+y)\sin (y+z)}{\sin x\sin z}\]                                 \[=\frac{BE}{AE}\times \frac{DC}{AD}\times \frac{AD}{BD}\times \frac{AE}{EC}\]                                 \[=\frac{2BD\times 2EC}{BD\times EC}=4.\]