10) In \( \triangle K H P, m \angle K=27^{\circ}, p=35 \mathrm{~m}, k=18 \) Find \( m \angle P \)

To find \( m \angle P \) in triangle \( KHP \) given \( m \angle K = 27^\circ \), \( p = 35 \, \text{m} \), and \( k = 18 \, \text{m} \), we can use the Law of Sines, which states: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] In this case, we can label the sides and angles as follows: - \( K \) corresponds to angle \( K \) and side \( k \) - \( H \) corresponds to angle \( H \) and side \( h \) - \( P \) corresponds to angle \( P \) and side \( p \) From the information given: - \( m \angle K = 27^\circ \) - \( k = 18 \, \text{m} \) (opposite angle \( H \)) - \( p = 35 \, \text{m} \) (opposite angle \( P \)) Using the Law of Sines, we can write: \[ \frac{p}{\sin P} = \frac{k}{\sin K} \] Substituting the known values: \[ \frac{35}{\sin P} = \frac{18}{\sin 27^\circ} \] Now, we need to calculate \( \sin 27^\circ \): \[ \sin 27^\circ \approx 0.4540 \] Now substituting this value into the equation: \[ \frac{35}{\sin P} = \frac{18}{0.4540} \] Calculating the right side: \[ \frac{18}{0.4540} \approx 39.64 \] Now we have: \[ \frac{35}{\sin P} = 39.64 \] Cross-multiplying gives: \[ 35 = 39.64 \cdot \sin P \] Now, solving for \( \sin P \): \[ \sin P = \frac{35}{39.64} \approx 0.884 \] Now we can find \( P \) by taking the inverse sine: \[ P \approx \sin^{-1}(0.884) \approx 61.0^\circ \] Thus, the measure of angle \( P \) is approximately: \[ \boxed{61^\circ} \]