Question

1. what is the measure of angle p?

Explanation:

Step1: Identify congruent triangles

Assuming $PQRS$ is a parallelogram, $\triangle PRS \cong \triangle RPQ$. Side $QS=9.3$ in, $RS=22$ in, $\angle S=35^\circ$. We use the Law of Sines on $\triangle PQS$ (or $\triangle PRS$) to find $\angle p_2$, then $\angle P = \angle p_1 + \angle p_2$, and $\angle p_1 = \angle s_1$.

Step2: Apply Law of Sines to $\triangle QRS$

Find $\angle r_2$:
$$\frac{\sin\angle r_2}{QS} = \frac{\sin 35^\circ}{RS}$$
$$\sin\angle r_2 = \frac{9.3 \times \sin35^\circ}{22}$$
$$\sin\angle r_2 = \frac{9.3 \times 0.5736}{22} \approx \frac{5.334}{22} \approx 0.2425$$
$$\angle r_2 \approx \arcsin(0.2425) \approx 14^\circ$$

Step3: Calculate $\angle s_1$

In $\triangle QRS$, sum of angles is $180^\circ$:
$$\angle s_1 = 180^\circ - 35^\circ - 14^\circ = 131^\circ$$
$\angle p_1 = \angle s_1 = 131^\circ$ (alternate interior angles in parallelogram)

Step4: Find $\angle p_2$ using Law of Sines on $\triangle PQS$

$\angle p_2 = \angle r_2 = 14^\circ$ (congruent triangles)

Step5: Calculate total $\angle P$

$$\angle P = \angle p_1 + \angle p_2 = 131^\circ + 14^\circ$$

Answer:

$145^\circ$

Note: This solution assumes $PQRS$ is a parallelogram, as the problem implies congruent triangles formed by diagonals, a key property of parallelograms.