In Delta PQR,PQ = PR,angle P = x + 10^{circ} | vedclass

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Question

(A) Given that in $\Delta PQR$,$PQ = PR$. Therefore,the angles opposite to these sides are equal,so $\angle Q = \angle R = 4x - 5^{\circ}$.By the angl

Vedclass · Accepted Answer

$30^{\circ}, 75^{\circ}, 75^{\circ}$

Vedclass · Answer

(A) Given that in $\Delta PQR$,$PQ = PR$. Therefore,the angles opposite to these sides are equal,so $\angle Q = \angle R = 4x - 5^{\circ}$.By the angle sum property of a triangle,$\angle P + \angle Q + \angle R = 180^{\circ}$.Substituting the given values: $(x + 10^{\circ}) + (4x - 5^{\circ}) + (4x - 5^{\circ}) = 180^{\circ}$.Combining like terms: $9x = 180^{\circ}$,which gives $x = 20^{\circ}$.Now,calculating the angles:$\angle P = x + 10^{\circ} = 20^{\circ} + 10^{\circ} = 30^{\circ}$.$\angle Q = 4x - 5^{\circ} = 4(20^{\circ}) - 5^{\circ} = 80^{\circ} - 5^{\circ} = 75^{\circ}$.Since $\angle Q = \angle R$,then $\angle R = 75^{\circ}$.Thus,the angles are $30^{\circ}, 75^{\circ}, 75^{\circ}$.

In $\Delta PQR$,$PQ = PR$,$\angle P = x + 10^{\circ}$ and $\angle Q = 4x - 5^{\circ}$. Find each of the angles of $\Delta PQR$.

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