In $\Delta ABC$, $A-B=60$ and $64\Delta^2=3abc(a+b+c)$, then find $\cos C$

Question

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

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

$$16\sin A\sin B\sin C =3(\sin A +\sin B+\sin C)$$

$$8(\cos (60)+\cos C)\sin C=3(2\cos \frac C2 ( \frac{\sqrt 3}{2})+\sin C)$$

$$8\sin C(\frac 12 +\cos C )=3(\sqrt 3 \cos \frac C2 +\sin C)$$

$$4\sin C (1+2\cos C)=3(\sqrt 3 \cos \frac C2 +\sin C)$$

$$4\sin C (4\cos ^2 \frac C2 -1)=3(\sqrt 3 \cos \frac C2 +\sin C)$$

$$8\sin \frac C2 (4\cos ^2 \frac C2 -1)=3(\sqrt 3 +2\sin \frac C2)$$

$$8(4-4\sin^2\frac C2 -1 )\sin \frac C2 =3(\sqrt 3 +2\sin \frac C2)$$

Solving the cubic

$$\sin \frac C2 =-\frac{\sqrt 3}{2}$$

So $\cos C =-\frac 12$

The given answer is $\frac 58$, but I don’t know what went wrong in my computation

Also I would like to avoid a cubic polynomial, since I used a calculator to solve this one.

Thanks

Answer 1

Continue with $$16\sin A \sin B \sin C= 3(\sin A+ \sin B +\sin C)$$

and simplify with $\sin A+ \sin B +\sin C= 4\cos\frac A2\cos\frac B2 \cos\frac C2$ $$32\sin\frac A2\sin\frac B2 \sin\frac C2=3 $$ or $$ 16 \left( \cos\frac{A-B}2-\sin\frac C2\right)\sin \frac C2=3 $$

which, with $A-B =60$, yields $\sin\frac C2=\frac{\sqrt3}4 $ and

$$\cos C = 1- 2\sin^2 \frac C2= \frac 58$$

Answer 2

\begin{align} \text{In $\triangle ABC$, if }\quad A-B&=60^\circ \tag{1}\label{1} ,\\ 64\Delta^2 &=3abc(a+b+c) \tag{2}\label{2} ,\\ \text{find }\cos C. \end{align}

Equation \eqref{2} in terms of semi-perimeter $\rho$, inradius $r$ and circumradius $R$ is

\begin{align} 64(\rho r)^2 &=3\cdot4\rho r R\cdot (2\rho) =24\rho^2 r R \tag{3}\label{3} ,\\ 64 r&=24 R \tag{4}\label{4} ,\\ \frac rR &=\frac{24}{64}=\frac38 \tag{5}\label{5} . \end{align}

Pairing \eqref{1} with $A+B+C=180^\circ$, we have

\begin{align} A &= 120^\circ-\tfrac12\,C ,\quad B = 60^\circ-\tfrac12\,C \tag{6}\label{6} . \end{align}

Recall that for the angles of triangle \begin{align} \cos A+\cos B+\cos C&=\frac rR+1 \tag{7}\label{7} ,\\ \text{so }\quad \cos(120^\circ-\tfrac12\,C)+ \cos(60^\circ-\tfrac12\,C)+ \cos C -\frac rR-1 &=0 \tag{8}\label{8} , \end{align}

\begin{align} \cos C+\sqrt3\sin\tfrac12C-\frac{11}8 &=0 \tag{9}\label{9} ,\\ \sin^2\tfrac12C -\tfrac{\sqrt3}2\, \sin\tfrac12C +\tfrac3{16} =(\sin\tfrac12C-\tfrac{\sqrt3}4)^2 &=0 \tag{10}\label{10} ,\\ \sin\tfrac12C &=\tfrac{\sqrt3}4 \tag{11}\label{11} ,\\ \cos C =1-2\sin^2\tfrac12C &=1-2\cdot(\tfrac{\sqrt3}4)^2 =\tfrac58 \tag{12}\label{12} , \end{align}

so the given answer is correct.