Question

In a triangle ABC, A – B = 60º and 64D2 = 3abc (a + b + c) {where a,b,c are sides of DABC opposite to vertices A,B,C respectively and D is area of the triangle}, then cosC is equal to-

Answer 1

Given:

        $A - B = 60^0$

        $64 \triangle^2 = 3abc(a+b+c)$

 Here $\triangle$ is area of triangle, a is side opposite to angle A and so on

    $64 \frac{a^2 b^2 c^2}{16R^2} = 3abc(a+b+C)$

     $16sin A sinB sinC = 3(sinA+sinB+sinC)$

simplifying with $sinA+sinB+sinC = 4cos\frac{A}{2} cos\frac{B}{2}cos\frac{c}{2}$

                                  $32sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2} = 3$

                        $16(cos\frac{A-B}{2} -sin\frac{C}{2}) sin \frac{C}{2} = 3$

with $A - B = 60,$

          $sin\frac{C}{2} = \frac{\sqrt{3}}{4}$ and

        $cos C = 1 -2sin^2\frac{C}{2} = \frac{5}{8}$

Answer 2

Given:

In triangle ABC, $A-B=60\textdegree$

and $64 \Delta^{2}=3abc(a+b+c)$

where here,

$\Delta$ is area of the triangle

$a,b,c$ are the sides opposite to angles $A,B,C$

We know:

$\Delta=\frac{abc}{4R}$ and $a=2R\times SinA$, $b=2R\times SinB$, $c=2R\times SinC$.

and $SinA+SinB+SinC=4Cos\frac{A}{2}\times Cos\frac{B}{2} \times Cos\frac{C}{2}$

Replacing the above in given, we get:

$64 \Delta^{2}=3abc(a+b+c)$

$64 (\frac{abc}{4R} )^{2}=3abc(a+b+c)$

$4\frac{abc}{R^{2} }=3(a+b+c)$

$4\times 2R\times SinA\times 2R\times SinB\times 2R\times SinC\times \frac{1}{R^{2} } =3(2R\times SinA+2R\times SinB+2R\times SinC)$

$16SinA\times SinB \times SinC=3(SinA+SinB+SinC)$

We also know that $SinA= 2Sin\frac{A}{2} \times Cos\frac{A}{2} ,SinB= 2Sin\frac{B}{2} \times Cos\frac{B}{2} , SinC= 2Sin\frac{C}{2} \times Cos\frac{C}{2}$

Therefore by placing in above equation:

$16SinA\times SinB \times SinC=3(SinA+SinB+SinC)$

$32Sin\frac{A}{2}\times Sin\frac{B}{2}\times Sin\frac{C}{2}=3$

$16(Cos\frac{A-B}{2} -Sin\frac{C}{2}) Sin\frac{C}{2}=3$

Replaceing $A-B=60\textdegree$, we get:

$Sin\frac{C}{2} =\frac{\sqrt{3} }{4}$

Therefore:

$CosC=1-2Sin^{2}\frac{C}{2}$

$CosC=1-2(\frac{\sqrt{3} }{4} )^{2}$

$CosC=1-2\times \frac{3}{16}$

$CosC=1-\frac{3}{8}$

$CosC=\frac{5}{8}$

The valueof $CosC=\frac{5}{8}$.