Question

lf the bisector of the angle $A$ makes an angle $\theta$ with $BC$, then Sin $\theta =$

• A. $Cos(\frac{B-C}{2})$
• B. $Sin(\frac{B-C}{2})$
• C. $Sin(B-\frac{A}{2})$
• D. $Sin(c-\frac{A}{2})$

Answer 1

The correct option is A $Cos(\frac{B-C}{2})$

$a=ksinA,b=ksinB,c=ksinC$

$\frac{sin\frac{A}{2}}{x}=\frac{sin\theta }{c}--\left(1\right),\frac{sin\frac{A}{2}}{y}=\frac{sin(\pi -\theta )}{b}--\left(2\right)$

$x=\frac{Csin\frac{A}{2}}{sin\theta },y=\frac{b-sin\frac{A}{2}}{sin\theta }$

$x+y=a$

$\frac{Csin\frac{A}{2}}{sin\theta }+\frac{bsin\frac{A}{2}}{sin\theta }=a$

$\frac{ksinCsin\frac{A}{2}}{sin\theta }+\frac{ksinBsin\frac{A}{2}}{sin\theta }=ksinA$

$\frac{sin\frac{A}{2}(sinC+sinB)}{sin\theta }=2cos\frac{A}{2}$

$2sin\left(\frac{C+B}{2}\right)cos\left(\frac{C-B}{2}\right)=sin\theta cos\frac{A}{2}$

$cos\left(\frac{A}{2}\right)cos\left(\frac{C-B}{2}\right)=sin\theta cos\frac{A}{2}$

$sin\theta =cos\left(\frac{B-C}{2}\right)$.