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If the perimeter of the triangle $A B C$ is $50 \mathrm{~cm}$, then $b \cos ^2 \frac{C}{2}+c \cos ^2 \frac{B}{2}=$
Options:
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Verified Answer
The correct answer is:
$25$
Since,
$b \cos ^2 \frac{C}{2}+c \cos ^2 \frac{B}{2}$
$\begin{aligned} & =\frac{1}{2}\left[b\left(2 \cos ^2 \frac{C}{2}\right)+c\left(2 \cos ^2 \frac{B}{2}\right)\right] \\ & =\frac{1}{2}[b(1+\cos C)+c(1+\cos B)]\end{aligned}$
$=\frac{1}{2}[b+c+(b \cos C+c \cos B)]=\frac{1}{2}[b+c+a]$
[from projection rule, $a=b \cos C+c \cos B$ ]
$=\frac{50}{2}=25 \mathrm{~cm}$
(it is given $a+b+c=50$ )
Hence,
option (b) is correct.
$b \cos ^2 \frac{C}{2}+c \cos ^2 \frac{B}{2}$
$\begin{aligned} & =\frac{1}{2}\left[b\left(2 \cos ^2 \frac{C}{2}\right)+c\left(2 \cos ^2 \frac{B}{2}\right)\right] \\ & =\frac{1}{2}[b(1+\cos C)+c(1+\cos B)]\end{aligned}$
$=\frac{1}{2}[b+c+(b \cos C+c \cos B)]=\frac{1}{2}[b+c+a]$
[from projection rule, $a=b \cos C+c \cos B$ ]
$=\frac{50}{2}=25 \mathrm{~cm}$
(it is given $a+b+c=50$ )
Hence,
option (b) is correct.
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