A) \[\sqrt{2}\]
B) \[\frac{\sqrt{3}+1}{2}\]
C) \[\frac{\sqrt{3}-1}{2}\]
D) 1
Correct Answer: A
Solution :
We know, \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] \[\therefore \] \[\cos B=\frac{{{3}^{2}}+{{4}^{2}}-{{5}^{2}}}{2\,(3)\,(4)}=\frac{9+16-25}{2\,(3)\,(4)}=0\] \[\Rightarrow \] \[B={{90}^{o}}\] \[\therefore \,\,\,\sin \frac{B}{2}+\cos \frac{B}{2}=\sin {{45}^{o}}+\cos {{45}^{o}}\] \[=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\]
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