ICSE BoardEnglish MediumSTD 9MATHEMATICSComplementary Angles2 Marks
Question
For triangle $\text{ABC},$ show that $:\sin \left(\frac{ A + B }{2}\right)=\cos \frac{ C }{2}$
✓
Answer
We know that for a triangle $\triangle ABC$
$\angle A + \angle B + \angle C = 180^\circ $
$\frac{\angle B +\angle A }{2}=90^{\circ}-\frac{\angle C }{2}$
$\sin \left(\frac{ A + B }{2}\right)=\sin \left(90^{\circ}-\frac{ C }{2}\right)$
$\sin \left(\frac{ A + B }{2}\right)=\cos \left(\frac{ C }{2}\right)$
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