In Delta ABC,m angle A + m angle C = m angle | vedclass

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(A) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^\circ$.Given $m \angle A + m \angle C = m \angle B$,substituting

Vedclass · Accepted Answer

$60$

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(A) In $\Delta ABC$,the sum of angles is $m \angle A + m \angle B + m \angle C = 180^\circ$.Given $m \angle A + m \angle C = m \angle B$,substituting this into the sum gives $m \angle B + m \angle B = 180^\circ$,so $2 m \angle B = 180^\circ$,which means $m \angle B = 90^\circ$.Since $\Delta ABC$ is a right-angled triangle at $B$,by the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.Let $AC = 17x$ and $AB = 15x$. Given $BC = 8$,we have $(17x)^2 = (15x)^2 + 8^2$.$289x^2 = 225x^2 + 64$.$64x^2 = 64$,so $x^2 = 1$,which gives $x = 1$.Thus,$AB = 15(1) = 15$ and $AC = 17(1) = 17$.The area of the right-angled triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes BC 	imes AB$.Area $= \frac{1}{2} 	imes 8 	imes 15 = 4 	imes 15 = 60$.

In $\Delta ABC$,$m \angle A + m \angle C = m \angle B$ and $AC : AB = 17 : 15$. If $BC = 8$,then the area of $\Delta ABC$ is.............

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