Question

m∠nqp = 45
m∠qpn = 53
m∠pnq = 82
rn =
pq =
nm = √

Explanation:

6 \), so \( \triangle MSN \) has \( SM = 16 \), \( SN = 16 \), so it's isoceles right triangle, so \( NM = 16\sqrt{2} \). Wait, the problem has \( NM = \_\_\sqrt{\_\_} \), so 16 and 2.

Step1: \( RN \)

From \( \triangle LSN \), right triangle with legs 12 and 16, hypotenuse 20. Notice that \( \triangle LSN \) and \( \triangle RNN \) (wait, \( RN \) is equal to \( LS = 12 \)? No, wait, the diagram has \( RN \) as the vertical segment, and \( \triangle PNR \) is right triangle with \( PN = 24\sqrt{2} \), \( \angle QPN = 53^{\circ} \), so \( RN = PN \times \sin(53^{\circ}) \). \( \sin(53^{\circ}) \approx \frac{4}{5} \), \( PN = 24\sqrt{2} \), so \( RN = 24\sqrt{2} \times \frac{4}{5} \)? No, this is confusing. Wait, the given \( \angle QPN = 53^{\circ} \), \( \angle NQP = 45^{\circ

Answer:

6 \), so \( \triangle MSN \) has \( SM = 16 \), \( SN = 16 \), so it's isoceles right triangle, so \( NM = 16\sqrt{2} \). Wait, the problem has \( NM = \_\_\sqrt{\_\_} \), so 16 and 2.

Step1: \( RN \)

From \( \triangle LSN \), right triangle with legs 12 and 16, hypotenuse 20. Notice that \( \triangle LSN \) and \( \triangle RNN \) (wait, \( RN \) is equal to \( LS = 12 \)? No, wait, the diagram has \( RN \) as the vertical segment, and \( \triangle PNR \) is right triangle with \( PN = 24\sqrt{2} \), \( \angle QPN = 53^{\circ} \), so \( RN = PN \times \sin(53^{\circ}) \). \( \sin(53^{\circ}) \approx \frac{4}{5} \), \( PN = 24\sqrt{2} \), so \( RN = 24\sqrt{2} \times \frac{4}{5} \)? No, this is confusing. Wait, the given \( \angle QPN = 53^{\circ} \), \( \angle NQP = 45^{\circ