A) equilateral triangle
B) acute angled triangle
C) obtuse angled triangle
D) right angled triangle
Correct Answer: B
Solution :
Given,\[x,\text{ }y,\text{ }z\]are in GP. \[\therefore \] \[{{y}^{2}}=xz\] and \[log\text{ }x-log\text{ }2y,\text{ }log\text{ }2y-log\text{ }3z\] and \[log\text{ }3z-log\text{ }x\]are in AP. \[\therefore \] \[2[log2y-log3z]\] \[=log\text{ }x-log\text{ }2y+log\text{ }3z-logx\] \[\Rightarrow \] \[2[log2y-log3z]=log3z-log2y\] \[\Rightarrow \] \[2log2y+log2y=log3z+2log3z\] \[\Rightarrow \] \[log2y=log3z\] \[\Rightarrow \] \[2y=3z\] Now, \[{{y}^{2}}=xz\] \[\Rightarrow \] \[{{y}^{2}}=x.\frac{2y}{3}\] \[\Rightarrow \] \[y=\frac{2x}{3}\] and \[z=\frac{2}{3}\times \frac{2x}{3}=\frac{4}{9}x\] Again now, \[x+y=x+\frac{2x}{3}=\frac{5x}{3}>z\] \[y+z=\frac{2x}{3}+\frac{4}{9}x=\frac{10x}{9}>x\] and \[z+x=\frac{4}{9}x+x=\frac{3}{9}x>y\] So,\[x,\text{ }y,\text{ }z\] are the sides of the triangle. \[\therefore \]\[\cos \theta =\frac{{{y}^{2}}+{{z}^{2}}-{{x}^{2}}}{2yz}=\frac{{{\left( \frac{2}{3}x \right)}^{2}}+{{\left( \frac{4}{9}x \right)}^{2}}-{{x}^{2}}}{2\left( \frac{2x}{3} \right)\left( \frac{4}{9}x \right)}\] \[=\frac{\frac{4{{x}^{2}}}{9}+\frac{16}{81}{{x}^{2}}-{{x}^{2}}}{\frac{16}{27}{{x}^{2}}}\] \[=\frac{\frac{-29}{81}}{\frac{16}{27}}\] \[=\frac{-29}{48}<0\] \[\therefore \]Triangle is a acute triangle.
You need to login to perform this action.
You will be redirected in
3 sec
