The area of triangle ABC is 84 sq. units. I | vedclass

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(A) Let $AB = c = 13$,$AC = b = 15$,and $BC = a$. The area of the triangle is given by $\Delta = 84$.Using the formula for the area of a triangle,$\De

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) Let $AB = c = 13$,$AC = b = 15$,and $BC = a$. The area of the triangle is given by $\Delta = 84$.Using the formula for the area of a triangle,$\Delta = \frac{1}{2} 	imes b 	imes c 	imes \sin(A)$,we have $84 = \frac{1}{2} 	imes 15 	imes 13 	imes \sin(A)$.$\sin(A) = \frac{84 	imes 2}{195} = \frac{168}{195} = \frac{56}{65}$.Since $\sin^2(A) + \cos^2(A) = 1$,$\cos(A) = \pm \sqrt{1 - (\frac{56}{65})^2} = \pm \sqrt{\frac{4225 - 3136}{4225}} = \pm \sqrt{\frac{1089}{4225}} = \pm \frac{33}{65}$.Using the Law of Cosines,$a^2 = b^2 + c^2 - 2bc \cos(A) = 15^2 + 13^2 - 2(15)(13) \cos(A) = 225 + 169 - 390 \cos(A) = 394 - 390 \cos(A)$.Case $1$: $\cos(A) = \frac{33}{65}$,$a^2 = 394 - 390(\frac{33}{65}) = 394 - 6(33) = 394 - 198 = 196$,so $a = 14$.Case $2$: $\cos(A) = -\frac{33}{65}$,$a^2 = 394 - 390(-\frac{33}{65}) = 394 + 198 = 592$,so $a = \sqrt{592} = 4\sqrt{37}$.Comparing with the options,$14$ is a valid value for $BC$.

The area of triangle $ABC$ is $84 \ sq. \ units$. If $AB = 13$ and $AC = 15$,then $BC$ can be ........ $units$.

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