Question

The sides a, b, c of △ABC are in G.P., where log a - log 2b, log 2b - log 3c, log 3c - log a are in A.P., then the △ABC is
(a) acute angled (b) obtuse angled (c) right angled (d) none of these

Answer 1

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$\huge\bold\red{{solution:-}}$

${b}^{2}=ac$

$2{(log 2b-log 3c)}$=${(log a-log b)}+{(log 3clog - log a)}$

$log\frac{(2b)}{(3c)}^{2}= log\frac{(3c)}{(2b)}$

$\frac{2b}{3c}^{2}=\frac{3c}{2b}$

➪${2b}^{3}={3c}^{3}$

${b}^{2}=a\frac{(2b)}{(3)}$

${b}=\frac{2a}{3}$

For b = 1 sides of square are ${3/2, 1 , 2/3}$

Now,

CosA =${4/9+1-9/4÷ 2×2/3×1 < 0}$

so, we get obtuse angled-------------------

Hence, Option (B) is the correct answer....

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