Question

The sides $a,b,c$ of $triangleABC$ are in $G.P$,.. where log $a-log2b$, $log2b-log3c$. $log3c-loga$, are in $A.P$. Then $△ABC$ is..

• A. Acute angled
• B. Obtuse angled
• C. right angled
• D. none of these

Answer 1

The correct option is B Obtuse angled

$\begin{array}{c}\begin{array}{c}{b}^2=ac\\ 2\left(\log 2b-\log 3c\right)=\left(\log a-\log b\right)+\left(\log 3c-\log a\right)\\ \log {\left(\frac{2b}{3c}\right)}^2=\log \left(\frac{3c}{2b}\right)\\ {\left(\frac{2b}{3c}\right)}^2=\frac{3c}{2b}\\ {\left(2b\right)}^3={\left(3c\right)}^3\\ \Rightarrow 2b=3c\\ b^2=a\left(\frac{2b}{3}\right)\\ b=\frac{2a}{3}\end{array}\\ For,b=1sidesoftriangleare\frac{3}{2},1,\frac{2}{3}\\ Nowfromthefigure\\ CosA=\frac{\frac{4}{9}+1-\frac{9}{4}}{2\times \frac{2}{3}\times 1}<0\\ so,wegetobtuseangled\\ Hence,optionBisthecorrectanswer.\end{array}$