With usual notations,in triangle ABC,the leng | vedclass

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(B) Given that angles $A, B, C$ are in $A$.$P$.,we have $2B = A + C$. Since $A+B+C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$.Using

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$24+\sqrt{6} 	ext{ cm}$

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(B) Given that angles $A, B, C$ are in $A$.$P$.,we have $2B = A + C$. Since $A+B+C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$.Using the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos B$.Let the sides be $a=10, c=9$ and $b$ be the third side. Then $b^2 = 10^2 + 9^2 - 2(10)(9) \cos 60^{\circ} = 100 + 81 - 180(0.5) = 181 - 90 = 91$.Thus $b = \sqrt{91}$. However,the problem implies the sides are $a, b, c$ where $B=60^{\circ}$.If $a=10, b=9$,then $9^2 = 10^2 + c^2 - 2(10)(c) \cos 60^{\circ} \implies 81 = 100 + c^2 - 10c \implies c^2 - 10c + 19 = 0$.Solving for $c$: $c = \frac{10 \pm \sqrt{100 - 76}}{2} = 5 \pm \sqrt{6}$.Case $1$: $c = 5 + \sqrt{6}$. Perimeter $= 10 + 9 + 5 + \sqrt{6} = 24 + \sqrt{6}$.Case $2$: $c = 5 - \sqrt{6}$. Perimeter $= 10 + 9 + 5 - \sqrt{6} = 24 - \sqrt{6}$.Given the options,$24+\sqrt{6}$ is a valid perimeter.

With usual notations,in $\triangle ABC$,the lengths of two sides are $10 \text{ cm}$ and $9 \text{ cm}$ respectively. If angles $A, B, C$ are in $A$.$P$.,then the perimeter of $\triangle ABC$ is:

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