Question

can anyone please solve this, I'm ready to give everything very difficult sum​

Answer 1

$\large\underline{\sf{Solution-4}}$

Given that, In triangle ABC, a = 13 cm, b = 14 cm and c = 15 cm

So, Semi-perimeter (s) of a triangle is given by

$\boxed{\rm{  \: \: s \: = \: \frac{a + b + c}{2} \: \: }}$

So, on substituting the values, we get

$\rm \: s \: = \: \frac{13 + 14 + 15}{2}$

$\rm \: s \: = \: \frac{42}{2}$

$\rm\implies \:s \: = \: 21 \: cm$

Now, we know that

$\rm \: sin \dfrac{A}{2} \: = \: \sqrt{ \dfrac{(s - b)(s - c)}{bc} }$

So, on substituting the values, we get

$\rm \: sin \dfrac{A}{2} \: = \: \sqrt{ \dfrac{(21 - 14)(21 - 15)}{14 \times 15} }$

$\rm \: sin \dfrac{A}{2} \: = \: \sqrt{ \dfrac{7 \times 6}{14 \times 15} }$

$\rm \: sin \dfrac{A}{2} \: = \: \sqrt{ \dfrac{1}{5} }$

$\rm\implies \:\rm \: sin \dfrac{A}{2} \: = \: \sqrt{ \dfrac{1}{5} }$

or

$\rm\implies \:\rm \: sin \dfrac{A}{2} \: = \: \dfrac{ \sqrt{5} }{5}$

$\large\underline{\sf{Solution-5}}$

Consider,

$\rm \: \dfrac{cosA}{a} + \dfrac{cosB}{b} + \dfrac{cosC}{c}$

We know, Cosine Law

$\boxed{\rm{  \:cosA = \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} \: \: }}$

$\boxed{\rm{  \:cosB = \frac{ {c}^{2} + {a}^{2} - {b}^{2} }{2ac} \: \: }}$

$\boxed{\rm{  \:cosC = \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab} \: \: }}$

So, on substituting these Identities, in given expression

$\rm \: = \: \dfrac{ {a}^{2} + {b}^{2} - {c}^{2} }{2abc} + \dfrac{ {c}^{2} + {a}^{2} - {b}^{2} }{2abc} + \dfrac{ {a}^{2} + {b}^{2} - {c}^{2} }{2abc}$

$\rm \: = \: \dfrac{ {a}^{2} + {b}^{2} - {c}^{2} + {c}^{2} + {a}^{2} - {b}^{2} + {a}^{2} + {b}^{2} - {c}^{2} }{2abc}$

$\rm \: = \: \dfrac{ {a}^{2}+{b}^{2} + {c}^{2}}{2abc}$

Hence,

$\rm\implies \:\frac{cosA}{a} + \frac{cosB}{b} + \frac{cosC}{c} = \: \dfrac{ {a}^{2}+{b}^{2} + {c}^{2}}{2abc}$

$\rule{190pt}{2pt}$

Additional Information :-

$\rm \: sin \dfrac{B}{2} \: = \: \sqrt{ \dfrac{(s - a)(s - c)}{ac} }$

$\rm \: sin \dfrac{C}{2} \: = \: \sqrt{ \dfrac{(s - a)(s - b)}{ab} }$

$\rm \: cos\dfrac{C}{2} \: = \: \sqrt{ \dfrac{s(s - c)}{ab} }$

$\rm \: cos\dfrac{B}{2} \: = \: \sqrt{ \dfrac{s(s - b)}{ac} }$

$\rm \: cos\dfrac{A}{2} \: = \: \sqrt{ \dfrac{s(s - a)}{ab} }$