The area of the triangle ABC is 10 sqrt{3} te | vedclass

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(A) Given: Area $= 10 \sqrt{3} 	ext{ cm}^2$,$\angle B = 60^{\circ}$,Perimeter $P = a + b + c = 20 	ext{ cm}$.Using the area formula: $	ext{Area} =

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Given: Area $= 10 \sqrt{3} 	ext{ cm}^2$,$\angle B = 60^{\circ}$,Perimeter $P = a + b + c = 20 	ext{ cm}$.Using the area formula: $	ext{Area} = \frac{1}{2} ac \sin B$.$10 \sqrt{3} = \frac{1}{2} ac \sin 60^{\circ} \implies 10 \sqrt{3} = \frac{1}{2} ac \left( \frac{\sqrt{3}}{2} ight) \implies ac = 40$.Using the Law of Cosines: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.$\cos 60^{\circ} = \frac{a^2 + c^2 - b^2}{2(40)} \implies \frac{1}{2} = \frac{(a+c)^2 - 2ac - b^2}{80}$.Since $a+c = 20 - b$,we have $40 = (20-b)^2 - 2(40) - b^2$.$40 = 400 - 40b + b^2 - 80 - b^2$.$40 = 320 - 40b$.$40b = 280 \implies b = 7$.Thus,$\ell(AC) = b = 7 	ext{ cm}$.

The area of the triangle $ABC$ is $10 \sqrt{3} \text{ cm}^2$,angle $B$ is $60^{\circ}$ and its perimeter is $20 \text{ cm}$,then $\ell(AC) = $ (in $\text{ cm}$)

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