In triangle ABC,if a, b, c are in arithmetic | vedclass

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(B) Given that $a, b, c$ are in arithmetic progression,so $2b = a + c$. Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} =

Vedclass · Accepted Answer

$2: 3$

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(B) Given that $a, b, c$ are in arithmetic progression,so $2b = a + c$. Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Thus,$a = k \sin A$,$b = k \sin B$,and $c = k \sin C$. Since $C = 2A$,we have $\sin C = \sin 2A = 2 \sin A \cos A$. From $2b = a + c$,we get $2 \sin B = \sin A + \sin C$. Using $B = 180^{\circ} - (A + C) = 180^{\circ} - 3A$,we have $\sin B = \sin 3A$. So,$2 \sin 3A = \sin A + \sin 2A$. $2(3 \sin A - 4 \sin^3 A) = \sin A + 2 \sin A \cos A$. Dividing by $\sin A$ (since $\sin A 
eq 0$): $6 - 8 \sin^2 A = 1 + 2 \cos A$. $6 - 8(1 - \cos^2 A) = 1 + 2 \cos A \Rightarrow 8 \cos^2 A - 2 \cos A - 3 = 0$. $(4 \cos A - 3)(2 \cos A + 1) = 0$. Since $A$ is an angle of a triangle,$\cos A = \frac{3}{4}$. Then $\frac{a}{c} = \frac{\sin A}{\sin C} = \frac{\sin A}{2 \sin A \cos A} = \frac{1}{2 \cos A} = \frac{1}{2(3/4)} = \frac{1}{3/2} = \frac{2}{3}$.

In $\triangle ABC$,if $a, b, c$ are in arithmetic progression and $C = 2A$,then $a: c =$

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