In $\triangle \mathrm{ABC}$, if the sides $a, b, c$ are in geometric progression and the largest angle…

In $\triangle \mathrm{ABC}$, if the sides $a, b, c$ are in geometric progression and the largest angle exceeds the smallest angle by $60^{\circ}$, then $\cos B$ is equal to
  1. $\frac{\sqrt{13}+1}{4}$
  2. $\frac{1-\sqrt{13}}{4}$
  3. $1$
  4. $\frac{\sqrt{13}-1}{4}$

Solution

In $\triangle \mathrm{ABC}$, sides $a, b$ and $c$ are in GP. $\therefore \quad b^2=a c \quad \ldots .(\mathrm{i})$ Given, largest angle exceeds the smallest angle by $60^{\circ}$. $\begin{aligned} & \mathrm{C}-\mathrm{A}=60^{\circ} \quad \ldots . \text { (ii) } \\ & \cos (\mathrm{C}-\mathrm{A})=\cos 60^{\circ} \\ & \cos \mathrm{C} \cos \mathrm{A}+\sin \mathrm{C} \sin \mathrm{A}=\frac{1}{2} \end{aligned}$ We know, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ $\begin{aligned} & \Rightarrow \quad 2 \cos \mathrm{C} \cos \mathrm{A}+2 \sin \mathrm{C} \sin \mathrm{A}=1 \\ & \Rightarrow \quad \cos (\mathrm{C}+\mathrm{A})+\cos (\mathrm{C}-\mathrm{A})+2 k^2 a c=1 \\ & \Rightarrow \quad \cos (\pi+\mathrm{B})+\cos 60^{\circ}+2 k^2 b^2=1 \quad\left[\therefore a c=b^2\right] \\ & \Rightarrow \quad-\cos \mathrm{B}+\frac{1}{2}+2 \sin ^2 \mathrm{~B}=1 \quad[\therefore b k=\sin \mathrm{B}] \\ & \Rightarrow \quad-\cos \mathrm{B}+1+4 \sin ^2 \mathrm{~B}=2 \\ & \Rightarrow \quad-2 \cos \mathrm{B}+1+4\left(1-\cos ^2 \mathrm{~B}\right)=2 \\ & \Rightarrow \quad 4 \cos ^2 \mathrm{~B}+2 \cos \mathrm{B}-3=0 \\ & \therefore \quad \cos \mathrm{B}=\frac{-2 \pm \sqrt{4+48}}{2 \times 4}=\frac{-2 \pm 2 \sqrt{13}}{8} \\ & =\frac{\sqrt{13}-1}{4} \text { or } \frac{-\sqrt{13}-1}{4} \end{aligned}$

Asked in: AP EAMCET 2016

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