In Delta PQR,PQ = PR and angle R = 40^{circ}, | vedclass

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(D) Given that in $\Delta PQR$,$PQ = PR$.Since the sides opposite to equal angles are equal,the angles opposite to equal sides are also equal.Therefor

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(D) Given that in $\Delta PQR$,$PQ = PR$.Since the sides opposite to equal angles are equal,the angles opposite to equal sides are also equal.Therefore,$\angle Q = \angle R$.Given $\angle R = 40^{\circ}$,so $\angle Q = 40^{\circ}$.The sum of all angles in a triangle is $180^{\circ}$.So,$\angle P + \angle Q + \angle R = 180^{\circ}$.Substituting the values: $\angle P + 40^{\circ} + 40^{\circ} = 180^{\circ}$.$\angle P + 80^{\circ} = 180^{\circ}$.$\angle P = 180^{\circ} - 80^{\circ} = 100^{\circ}$.

In $\Delta PQR$,$PQ = PR$ and $\angle R = 40^{\circ}$,then $\angle P = \dots$ (in $^{\circ}$)

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