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(A) Let the sides of the triangle be $a, b, c$ in $A.P.$ such that $a < b < c$. Then $2b = a + c$.Let the angles be $A, B, C$ opposite to sides $a, b,

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$4 : 5 : 6$

Vedclass · Answer

(A) Let the sides of the triangle be $a, b, c$ in $A.P.$ such that $a Let the angles be $A, B, C$ opposite to sides $a, b, c$ respectively. Given $C = 2A$.Since $a In any triangle,$A + B + C = 180^{\circ}$. Since $C = 2A$,$B = 180^{\circ} - 3A$.By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.From $2b = a + c$,we get $2\sin B = \sin A + \sin C$.$2\sin(180^{\circ} - 3A) = \sin A + \sin 2A$.$2\sin 3A = \sin A + 2\sin A \cos A$.$2(3\sin A - 4\sin^3 A) = \sin A(1 + 2\cos A)$.Since $\sin A 
eq 0$,$6 - 8\sin^2 A = 1 + 2\cos A$.$6 - 8(1 - \cos^2 A) = 1 + 2\cos A$.$8\cos^2 A - 2\cos A - 3 = 0$.$(4\cos A + 3)(2\cos A - 1) = 0$.Since $A$ is an angle of a triangle,$\cos A = \frac{3}{4}$ (as $\cos A = -\frac{3}{4}$ implies an obtuse angle,but $C=2A$ would exceed $180^{\circ}$).The ratio of sides is $\sin A : \sin B : \sin C = \sin A : \sin 3A : \sin 2A$.$= 1 : (3 - 4\sin^2 A) : 2\cos A$.$= 1 : (3 - 4(1 - \frac{9}{16})) : 2(\frac{3}{4}) = 1 : (3 - 4(\frac{7}{16})) : \frac{3}{2} = 1 : \frac{5}{4} : \frac{6}{4} = 4 : 5 : 6$.

If the lengths of the sides of a triangle are in $A.P.$ and the greatest angle is double the smallest,then the ratio of the lengths of the sides of this triangle is:

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