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

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(A) Given that $m \angle A + m \angle B = m \angle C$. Since the sum of angles in a triangle is $180^{\circ}$,we have $m \angle A + m \angle B + m \an

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(A) Given that $m \angle A + m \angle B = m \angle C$. Since the sum of angles in a triangle is $180^{\circ}$,we have $m \angle A + m \angle B + m \angle C = 180^{\circ}$. Substituting $m \angle A + m \angle B = m \angle C$,we get $m \angle C + m \angle C = 180^{\circ}$,which implies $2m \angle C = 180^{\circ}$,so $m \angle C = 90^{\circ}$. Thus,$\Delta ABC$ is a right-angled triangle with the hypotenuse $AB = 17.5$. Given $AC : BC = 3 : 4$,let $AC = 3x$ and $BC = 4x$. By the Pythagorean theorem,$AC^2 + BC^2 = AB^2$. $(3x)^2 + (4x)^2 = (17.5)^2$. $9x^2 + 16x^2 = 306.25$. $25x^2 = 306.25$. $x^2 = 12.25$,so $x = 3.5$. Thus,$AC = 3(3.5) = 10.5$ and $BC = 4(3.5) = 14$. The perimeter of $\Delta ABC = AC + BC + AB = 10.5 + 14 + 17.5 = 42$.

In $\Delta ABC$,$m \angle A + m \angle B = m \angle C$. If $AC : BC = 3 : 4$ and $AB = 17.5$,find the perimeter of $\Delta ABC$.

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