Trigonometric Equations Question 136
Question: 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}= $
Options:
A) 1
B) 2
C) 4
D) None of these
Show Answer
Answer:
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 $ .
BETA
