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\[\frac{\sin (x+y)\sin (y+z)}{\sin x\sin z}\] \[=\frac{BE}{AE}\times \frac{DC}{AD}\times \frac{AD}{BD}\times \frac{AE}{EC}=\frac{2BD\times 2EC}{BD\times EC}=4\].You need to login to perform this action.
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