Question In \( \triangle \mathrm{PQR}, q=420 \) inches, \( p=400 \) inches and \( \angle \mathrm{P}=64^{\circ} \). Find all possible values of \( \angle \mathrm{Q} \), to the nearest degree.

To find the possible values of \( \angle Q \) in triangle \( \triangle PQR \), we can use the Law of Sines, which states: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] Here, we have: - \( p = 400 \) inches (opposite angle \( P \)) - \( q = 420 \) inches (opposite angle \( Q \)) - \( \angle P = 64^\circ \) Using the Law of Sines: \[ \frac{p}{\sin P} = \frac{q}{\sin Q} \] Substituting the values we have: \[ \frac{400}{\sin 64^\circ} = \frac{420}{\sin Q} \] Now, solving for \( \sin Q \): \[ \sin Q = \frac{420 \cdot \sin 64^\circ}{400} \] Calculating this gives: \[ \sin Q \approx \frac{420 \cdot 0.4384}{400} \approx 0.4595 \] Next, we find \( \angle Q \) by taking the inverse sine: \[ \angle Q \approx \arcsin(0.4595) \approx 27^\circ \] However, because of the sine function's property, we have another possible angle which is: \[ \angle Q' = 180^\circ - \angle Q \approx 180^\circ - 27^\circ = 153^\circ \] Thus the two possible values for \( \angle Q \) rounded to the nearest degree are approximately \( 27^\circ \) and \( 153^\circ \).