In Delta ABC,mangle A + mangle C = mangle B. | vedclass

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(A) Given that in $\Delta ABC$,$m\angle A + m\angle C = m\angle B$.We know that the sum of all angles in a triangle is $180^{\circ}$,so $m\angle A + m

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) Given that in $\Delta ABC$,$m\angle A + m\angle C = m\angle B$.We know that the sum of all angles in a triangle is $180^{\circ}$,so $m\angle A + m\angle B + m\angle C = 180^{\circ}$.Substituting $m\angle A + m\angle C = m\angle B$ into the equation,we get $m\angle B + m\angle B = 180^{\circ}$,which implies $2m\angle B = 180^{\circ}$,so $m\angle B = 90^{\circ}$.Since $\angle B = 90^{\circ}$,$\Delta ABC$ is a right-angled triangle with $AC$ as the hypotenuse.By the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.Substituting the given values,$AC^2 = 7^2 + 24^2 = 49 + 576 = 625$.Therefore,$AC = \sqrt{625} = 25$.

In $\Delta ABC$,$m\angle A + m\angle C = m\angle B$. If $AB = 7$ and $BC = 24$,then $AC = \ldots$

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