Question

I want the solution of both sum​

Answer 1

$\large\underline{\sf{Solution-2}}$

Given that, in triangle ABC

$\rm \: a \: cosA \: = \: b \: cosB$

We know, By sine law, we have

$\boxed{\rm{  \: \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC} = k \: }}$

So, from this we have

$\rm \: a = k \: sinA$

$\rm \: b = k \: sinB$

So, on substituting these values, we get

$\rm \: k \: sinA \: cosA \: = \: k \: sinB \: cosB$

$\rm \: \: sinA \: cosA \: = \: \: sinB \: cosB$

On multiply and divide by 2, we get

$\rm \: 2\: sinA \: cosA \: = \: 2\: sinB \: cosB$

$\rm \: sin2A = sin2B$

$\rm \: sin2A - sin2B = 0 \:$

We know

$\boxed{\rm{  \:\rm \:sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg] \: \: }}$

So, using this result, we get

$\rm \: 2cos(A + B) \: sin(A - B) = 0$

$\rm \: cos(A + B) = 0\: \: or \: \: sin(A - B) = 0$

$\rm\implies \:A + B = \dfrac{\pi}{2} \: \: or \: \: A - B = 0$

$\rm\implies \:A + B = \dfrac{\pi}{2} \: \: or \: \: A = B$

$\rm\implies \: \triangle \: ABC \: is \: isosceles \: or \: right \: angled \: triangle$

$\large\underline{\sf{Solution-3}}$

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

We know,

Semi-perimeter (s) of a triangle with sides a, b and c, is given by

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

So, on substituting the values, we get

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

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

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

Now, we know

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

So, on substituting the values, we get

$\rm \: tan\dfrac{B}{2} = \sqrt{ \dfrac{(21 - 13)(21 - 15)}{21(21 - 14)} }$

$\rm \: tan\dfrac{B}{2} = \sqrt{ \dfrac{(8)(6)}{21(7)} }$

$\rm \: tan\dfrac{B}{2} = \sqrt{ \dfrac{(8)(2)}{(7)(7)} }$

$\rm\implies \:\rm \: tan\dfrac{B}{2} = \dfrac{4}{7}$

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

Additional Information :-

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

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

Answer 2

Step-by-step explanation:

$\boxed{ \underline{\underline{\text{Question 3}}}}$

$\\ \\ \tt \: s =\frac{a+b+c}{2}=\frac{13+14+15}{2}=21$

$\begin{array}{l} \tt( s - a )=21-13=8 \\ \\ \tt ( s - b )=21-14=7 \\ \\ \tt( s - c )=21-15=6 \end{array}$

$\\ \\ \color{orangered} \tt\tan \frac{B}{2}=\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}=\frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}}=\frac{1}{2}$