A) Acute angled
B) Obtuse angled
C) Right angled
D) Equilateral
Correct Answer: B
Solution :
As given \[{{b}^{2}}=ac\] and \[2(\log 2b-\log 3c)=\log a-\log 2b+\log 3c-\log a\] \[\Rightarrow \]\[{{b}^{2}}=ac\] and \[2b=3c\]\[\Rightarrow \]\[b=2a/3\]and\[c=4a/9\] Since \[a+b=\frac{5a}{3}>c,\ b+c=\frac{10a}{9},>a,\ c+a=\frac{13a}{9}>b\] It implies that \[a,\ b,\ c\] form \[a\] triangle with \[a\] as the greatest side. Now, let us find the greatest angle \[A\] of \[\Delta ABC\] by using the cosine formula. \[\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=-\frac{29}{48}<0\] \[\therefore \] The angle \[A\] is obtuse.
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