If 7 and 8 are the lengths of two sides of a | vedclass

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(B) Let the angles of the triangle be $A - d, A, A + d$. Since the sum of angles is $180^{\circ}$,we have $3A = 180^{\circ}$,so $A = 60^{\circ}$.Thus,

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$21$

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(B) Let the angles of the triangle be $A - d, A, A + d$. Since the sum of angles is $180^{\circ}$,we have $3A = 180^{\circ}$,so $A = 60^{\circ}$.Thus,the angles are $60^{\circ} - d, 60^{\circ}, 60^{\circ} + d$.The sides opposite to these angles are $a, b, c$. Given two sides are $7$ and $8$. Let the sides be $a, 7, 8$ where $a$ is the smallest side.Using the Law of Cosines for the angle $60^{\circ}$:Case $1$: If the angle $60^{\circ}$ is between sides $7$ and $8$,then $a^2 = 7^2 + 8^2 - 2(7)(8) \cos 60^{\circ} = 49 + 64 - 56 = 57$. This gives $a = \sqrt{57} \approx 7.55$,which is not smaller than $7$.Case $2$: If the angle $60^{\circ}$ is opposite to side $7$,then $8^2 = a^2 + 7^2 - 2(a)(7) \cos 60^{\circ}$ $\Rightarrow 64 = a^2 + 49 - 7a$ $\Rightarrow a^2 - 7a - 15 = 0$. This yields $a = \frac{7 \pm \sqrt{49 + 60}}{2} = \frac{7 \pm \sqrt{109}}{2}$. Only the positive value is valid,$a \approx 8.72 > 7$.Case $3$: If the angle $60^{\circ}$ is opposite to side $8$,then $7^2 = a^2 + 8^2 - 2(a)(8) \cos 60^{\circ}$ $\Rightarrow 49 = a^2 + 64 - 8a$ $\Rightarrow a^2 - 8a + 15 = 0$.Solving $a^2 - 8a + 15 = 0$ gives $(a - 3)(a - 5) = 0$,so $a = 3$ or $a = 5$.Since $a$ is the smallest side,both $3 Thus,$a_1 = 3$ and $a_2 = 5$.Finally,$2 a_1 + 3 a_2 = 2(3) + 3(5) = 6 + 15 = 21$.

If $7$ and $8$ are the lengths of two sides of a triangle and '$a$' is the length of its smallest side. The angles of the triangle are in $AP$ and '$a$' has two values $a_1$ and $a_2$ satisfying this condition. If $a_1 < a_2$ then $2 a_1 + 3 a_2 =$

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