In Delta PQR,m angle P = m angle Q + m angle | vedclass

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(C) In $\Delta PQR$,we are given $m \angle P = m \angle Q + m \angle R$.We know that the sum of all angles in a triangle is $180^{\circ}$,so $m \angle

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$60$

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(C) In $\Delta PQR$,we are given $m \angle P = m \angle Q + m \angle R$.We know that the sum of all angles in a triangle is $180^{\circ}$,so $m \angle P + m \angle Q + m \angle R = 180^{\circ}$.Substituting $m \angle Q + m \angle R$ with $m \angle P$,we get $m \angle P + m \angle P = 180^{\circ}$,which implies $2 \cdot m \angle P = 180^{\circ}$,so $m \angle P = 90^{\circ}$.Since $\Delta PQR$ is a right-angled triangle with the right angle at $P$,$QR$ is the hypotenuse.By the Pythagorean theorem,$QR^2 = PQ^2 + PR^2$.Substituting the given values,$25^2 = 20^2 + PR^2$.$625 = 400 + PR^2$,so $PR^2 = 625 - 400 = 225$.Thus,$PR = \sqrt{225} = 15$.The perimeter of $\Delta PQR = PQ + QR + PR = 20 + 25 + 15 = 60$.

In $\Delta PQR$,$m \angle P = m \angle Q + m \angle R$. If $PQ = 20$ and $QR = 25$,then the perimeter of $\Delta PQR$ is............

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