In a Delta ABC,if the sides are a = 3, b = 5, | vedclass

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(A) Given sides are $a = 3, b = 5, c = 4$. First,calculate the semi-perimeter $s$: $s = \frac{a + b + c}{2} = \frac{3 + 5 + 4}{2} = \frac{12}{2} = 6$.

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(A) Given sides are $a = 3, b = 5, c = 4$. First,calculate the semi-perimeter $s$: $s = \frac{a + b + c}{2} = \frac{3 + 5 + 4}{2} = \frac{12}{2} = 6$. Using the half-angle formulas for a triangle: $\sin \frac{B}{2} = \sqrt{\frac{(s - a)(s - c)}{ac}} = \sqrt{\frac{(6 - 3)(6 - 4)}{3 	imes 4}} = \sqrt{\frac{3 	imes 2}{12}} = \sqrt{\frac{6}{12}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$. $\cos \frac{B}{2} = \sqrt{\frac{s(s - b)}{ac}} = \sqrt{\frac{6(6 - 5)}{3 	imes 4}} = \sqrt{\frac{6 	imes 1}{12}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$. Therefore,$\sin \frac{B}{2} + \cos \frac{B}{2} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

In a $\Delta ABC$,if the sides are $a = 3, b = 5$,and $c = 4$,then $\sin \frac{B}{2} + \cos \frac{B}{2}$ is equal to

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