Question

In the diagram below, quadrilateral NOPQ is inscribed in circle R. Find themeasure of ZP.N70°o94RQ

Answer 1

Answer:

P = 110 degrees

Explanation:

We were given the following information:

A quadrilateral NOPQ is inscribed in a circle. This makes the quadrilateral a cyclic quadrilateral

$\begin{gathered} \angle N=70^{\circ} \\ \angle O=94^{\circ} \end{gathered}$

Angle P is supplementary to Angle N; both angles P & N sum up to 180 degrees:

$\begin{gathered} \angle P+\angle N=180^{\circ} \\ \angle N=70^{\circ} \\ \angle P+70^{\circ}=180^{\circ} \\ \text{Subtract ''}70^{\circ}\text{'' from both sides, we have:} \\ \angle P=(180-70)^{\circ} \\ \angle P=110^{\circ} \\ \\ \therefore\angle P=110^{\circ} \end{gathered}$

Therefore, the angle at P equals 110 degrees