If the perimeter of the triangle ABC is 50 te | vedclass

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Question

(B) Given the expression: $b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2}$ Using the identity $2 \cos^2 	heta = 1 + \cos(2	heta)$,we have: $= \frac{1}{

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) Given the expression: $b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2}$ Using the identity $2 \cos^2 	heta = 1 + \cos(2	heta)$,we have: $= \frac{1}{2} [b(1 + \cos C) + c(1 + \cos B)]$ $= \frac{1}{2} [b + c + b \cos C + c \cos B]$ By the projection rule,$a = b \cos C + c \cos B$. Substituting this into the expression: $= \frac{1}{2} [b + c + a] = \frac{1}{2} (a + b + c)$ Since the perimeter $a + b + c = 50 	ext{ cm}$: $= \frac{50}{2} = 25 	ext{ cm}$. Thus,option $B$ is correct.

If the perimeter of the triangle $ABC$ is $50 \text{ cm}$,then $b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} =$

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