Question

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 $\frac{sin(x+y)sin(y+z)}{sinxsinz}$

• A. 1
• B. 2
• C. 4
• D. 3

Answer 1

The correct option is C

4

$From\Delta ADC,\frac{sin(y+z)}{DC}=\frac{sinC}{AD}\text{Applying sine rule}From\Delta ABD,\frac{sinx}{BD}=\frac{sinB}{AD}$

$From\Delta AEC,\frac{sinz}{EC}=\frac{sinB}{AE}\therefore \frac{sin(x+y)sin(y+z)}{sinxsinz}=\frac{BE}{BD}.\frac{DC}{EC}=2\times 2=4$