Answer:
See Explanation
Step-by-step explanation:
[tex]A+B+C=180° \\ \\ A+B=180° -C \\ \\ dividing \: throughout \: by \: 2 \\ \\ \frac{A+B}{2}= \frac{180° -C}{2} \\ \\ \frac{A+B}{2}= \frac{180 \degree}{2} - \frac{C}{2} \\ \\ \frac{A+B}{2}= 90 \degree - \frac{C}{2} \\ \\ taking \: \sin \: ratio \: on \: both \: sides \\ \\ sin \bigg(\frac{A+B}{2} \bigg)= \sin \bigg(90 \degree - \frac{C}{2} \bigg) \\ \\sin \bigg(\frac{A+B}{2} \bigg)= \cos \bigg( \frac{C}{2} \bigg) \\ (\because \sin(90 \degree - \theta) = \cos \theta )\\ \\ thus \: proved[/tex]
