Question

If the sides $a,b,c$ of a triangle are in G.P and largest angle exceeds the smallest by ${60}^0$, then $\cos B$ is equal to :

Answer 1

$a,b,c$ are in G.P.
$\Rightarrow b^2=ac$
Using sine rule, we have
$\Rightarrow {\sin }^{2}B=\sin A\sin C$
Multiply both sides by $2$ we get
$\Rightarrow 2{\sin }^{2}B=2\sin A\sin C$
$\Rightarrow 2(1-{\cos }^{2}B)=\cos (A-C)-\cos (A+C)$
Given:largest angle exceeds the smallest by ${60}^0$
$\Rightarrow A={60}^0+C$ or $A-C={60}^0$
$\Rightarrow 2-2{\cos }^{2}B=\cos {60}^0-\cos ({180}^0-B)$
$\Rightarrow 2-2{\cos }^{2}B=\frac{1}{2}+\cos B$
$\Rightarrow 2{\cos }^{2}B+\cos B-2+\frac{1}{2}=0$
$\Rightarrow 4{\cos }^{2}B+2\cos B-3=0$
$\Rightarrow \cos B=\frac{-2\pm \sqrt{4+48}}{8}$
But $\left|\cos B\right|<1$ so that
$\cos B=\frac{\sqrt{13}-1}{4}$