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

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(C) Given that $a, b, c$ are in arithmetic progression,we have $2b = a + c$. Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.

Vedclass · Accepted Answer

$2 : 9 : 12$

Vedclass · Answer

(C) Given that $a, b, c$ are in arithmetic progression,we have $2b = a + c$. Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$. Substituting these,$2 \sin B = \sin A + \sin C$. Given $A = 2C$,we have $2 \sin B = \sin 2C + \sin C = 2 \sin C \cos C + \sin C = \sin C (2 \cos C + 1)$. Since $B = 180^{\circ} - (A + C) = 180^{\circ} - 3C$,we have $\sin B = \sin 3C = 3 \sin C - 4 \sin^3 C$. Substituting this,$2(3 \sin C - 4 \sin^3 C) = \sin C (2 \cos C + 1)$. Dividing by $\sin C$ (as $\sin C 
eq 0$),$6 - 8 \sin^2 C = 2 \cos C + 1$. Using $\sin^2 C = 1 - \cos^2 C$,we get $6 - 8(1 - \cos^2 C) = 2 \cos C + 1$,which simplifies to $8 \cos^2 C - 2 \cos C - 3 = 0$. Solving this quadratic equation,$(4 \cos C + 3)(2 \cos C - 1) = 0$. Since $C$ is an angle of a triangle,$\cos C = 1/2$,so $C = 60^{\circ}$. Then $A = 2C = 120^{\circ}$ and $B = 180^{\circ} - 180^{\circ} = 0^{\circ}$,which is impossible. Wait,re-evaluating: $a, b, c$ are in $AP$ implies $a+c=2b$. By sine rule,$\sin A + \sin C = 2 \sin B$. Using $A=2C$ and $B=180-3C$,$\sin 2C + \sin C = 2 \sin 3C$. $2 \sin C \cos C + \sin C = 2(3 \sin C - 4 \sin^3 C)$. $2 \cos C + 1 = 6 - 8 \sin^2 C = 6 - 8(1 - \cos^2 C) = 8 \cos^2 C - 2$. $8 \cos^2 C - 2 \cos C - 3 = 0$. Roots are $\cos C = \frac{2 \pm \sqrt{4 + 96}}{16} = \frac{2 \pm 10}{16}$. $\cos C = 3/4$ or $\cos C = -1/2$. Since $A=2C Then $\cos A = \cos 2C = 2 \cos^2 C - 1 = 2(9/16) - 1 = 1/8$. $\sin C = \sqrt{1 - 9/16} = \sqrt{7}/4$. $\sin A = \sin 2C = 2 \sin C \cos C = 2(\sqrt{7}/4)(3/4) = 3\sqrt{7}/8$. $\sin B = \sin(180 - 3C) = \sin 3C = 3 \sin C - 4 \sin^3 C = \sin C (3 - 4 \sin^2 C) = \frac{\sqrt{7}}{4} (3 - 4(7/16)) = \frac{\sqrt{7}}{4} (3 - 7/4) = \frac{\sqrt{7}}{4} (5/4) = 5\sqrt{7}/16$. $\cos B = \cos(180 - 3C) = -\cos 3C = -(4 \cos^3 C - 3 \cos C) = -\cos C (4 \cos^2 C - 3) = -\frac{3}{4} (4(9/16) - 3) = -\frac{3}{4} (9/4 - 3) = -\frac{3}{4} (-3/4) = 9/16$. Thus $\cos A : \cos B : \cos C = 1/8 : 9/16 : 3/4 = 2/16 : 9/16 : 12/16 = 2 : 9 : 12$.

In a triangle $ABC$,if $a, b, c$ are in arithmetic progression and the angle $A$ is twice the angle $C$,then $\cos A : \cos B : \cos C =$

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