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If the perimeter of a triangle $\mathrm{ABC}$ is $30 \mathrm{~cm}$, then what is the value of $a \cos ^{2}(\mathrm{C} / 2)+c \cos ^{2}(\mathrm{~A} / 2) ?$
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The correct answer is:
$15 \mathrm{~cm}$
We know from properties of triangle that
$\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{b c}} \Rightarrow \cos ^{2} \frac{A}{2}=\frac{s(s-a)}{b c}$
and $\cos \frac{\mathrm{C}}{2}=\sqrt{\frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}} \Rightarrow \cos ^{2} \frac{\mathrm{C}}{2}=\frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}$
So, $a \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{A}{2}$
$=\mathrm{a}\left(\frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}\right)+\mathrm{c}\left(\frac{\mathrm{s}(\mathrm{s}-\mathrm{a})}{\mathrm{bc}}\right)=\frac{\mathrm{s}(\mathrm{s}-\mathrm{c}+\mathrm{s}-\mathrm{a})}{\mathrm{b}}$
$=\frac{s(2 s-a-c)}{b}=\frac{s(a+b+c-a-c)}{b}=\frac{s \cdot b}{b}=s$
$=\frac{30}{2}=15 \mathrm{~cm}$ [given that $2 \mathrm{~s}=30$ ]
$\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{b c}} \Rightarrow \cos ^{2} \frac{A}{2}=\frac{s(s-a)}{b c}$
and $\cos \frac{\mathrm{C}}{2}=\sqrt{\frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}} \Rightarrow \cos ^{2} \frac{\mathrm{C}}{2}=\frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}$
So, $a \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{A}{2}$
$=\mathrm{a}\left(\frac{\mathrm{s}(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}\right)+\mathrm{c}\left(\frac{\mathrm{s}(\mathrm{s}-\mathrm{a})}{\mathrm{bc}}\right)=\frac{\mathrm{s}(\mathrm{s}-\mathrm{c}+\mathrm{s}-\mathrm{a})}{\mathrm{b}}$
$=\frac{s(2 s-a-c)}{b}=\frac{s(a+b+c-a-c)}{b}=\frac{s \cdot b}{b}=s$
$=\frac{30}{2}=15 \mathrm{~cm}$ [given that $2 \mathrm{~s}=30$ ]
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