Question

In a ∆ABC if a=13, b=14 and c=15

Find:
i)sinA/2
ii)cosA/2
iii)CosC​

Answer 1

$\huge{\underline{\sf{\red{Answer}}}}$

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Given:

•a=13

•b=14

•c=15

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$\huge{\underline{\underline{\green{\tt{Formula}}}}}$

Semi-Perimeter = $\dfrac{a+b+c}{2}$

$\implies Sin\dfrac{A}{2} = \sqrt{\dfrac{(s-b)(s-c)}{bc}} \\ \implies Cos \dfrac{A}{2} =\sqrt{\dfrac{s(s-a))}{bc}} \\ \implies CosC= \dfrac{a^2 + b^2 -c^2}{2ab}$

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SoluTion:

S=Semi-Perimeter

$\implies\dfrac{a+b+c}{2} \\ \implies\dfrac{13+14+15}{2} \\ \implies \dfrac{42}{2} \\ \huge{\boxed{s\leadsto 21}}$

i)Sin$\dfrac{A}{2}$

by putting Values in the formula we get,

$\sqrt{\dfrac{(21-14)(21-15)}{14\times 15}} \\ \implies \sqrt{\dfrac{\cancel{7} \times \cancel{6}}{\cancel{14} \times \cancel{15}}} \\ \implies \sqrt{\dfrac{1}{5}} \\ \huge{\boxed{\dfrac{1}{\sqrt{5}}}}$

ii)Cos$\dfrac{A}{2}$

Similarly by putting Values in formula

$\sqrt{\dfrac{21(21-13)}{14\times 15}} \\ \implies \sqrt{\dfrac{\cancel{21} \times \cancel{8}}{\cancel{14} \times \cancel{15}}} \\ \implies{\huge{\boxed{\dfrac{2}{\sqrt{5}}}}}$

iii)CosC

$\dfrac{13^2 +14^2-13^2}{2\times 13 \times 14} \\ \implies {\dfrac{169+196-225}{28 \times 13 }} \\ \implies{\dfrac{\cancel{140}}{\cancel{28}\times{13}}} \\ \implies{\huge{\boxed{\dfrac{5}{13}}}}$

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