Question

prove that aSin(A/2+B)=(b+c)SinA/2

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

In triangle ABC,we have AB=c, AC=b and BC=a.

Then, we have to prove that $asin(\frac{A}{2}+B)=(b+c)sin\frac{A}{2}$.

Now, using the sine formulae,

⇒$\frac{b+c}{a}= \frac{KsinB+KsinC}{KsinA}$

⇒$\frac{b+c}{a}=\frac{K(sinB+sinC) }{KsinA}$

⇒$\frac{b+c}{a}=\frac{2sin(\frac{B+C}{2})cos(\frac{B-C}{2}) }{2sin\frac{A}{2}cos\frac{A}{2}  }$

⇒$\frac{b+c}{a}=\frac{sin\frac{{\pi}-A}{2}cos\frac{B-C}{2}  }{sin\frac{A}{2}cos\frac{A}{2}  }$ (using A+B+C=180°)

⇒$\frac{b+c}{a}=\frac{sin(\frac{{\pi}}{2}-\frac{A}{2})cos(\frac{B-C}{2})  }{sin\frac{A}{2}cos\frac{A}{2} }$

Now, $sin(\frac{{\pi}}{2}-\frac{A}{2})=cos\frac{A}{2}$], therefore,

⇒$\frac{b+c}{a}=\frac{cos(\frac{B-C}{2}) }{sin\frac{A}{2} }$

Putting C= π-A-B, we have

⇒$\frac{b+c}{a}=\frac{cos(\frac{B-({\pi}-A-B)}{2})}{sin\frac{A}{2} }$

⇒$\frac{b+c}{a}=\frac{cos(\frac{{\pi}}{2}-(\frac{A}{2}+B))  }{sin\frac{A}{2} }$

⇒$\frac{b+c}{a}=\frac{sin(\frac{A}{2}+B) }{sin\frac{A}{2} }$

⇒$asin(\frac{A}{2}+B)= (b+c)sin\frac{A}{2}$

Hence proved.