S0S1=15 (miles)
∠L1S0N0=∠L2S0N0=40°
∠L2S0P=50°
∠N0S0S1=135°
∠L2S0S1=135°−40°=95°
∠L1S1N1=10°
∠N1S1L2=360°−345°=15°
∆TPS1:∠TPS1=90°−15°=75°=∠L2PS0
∆PL2S0:∠PL2S0=180°−∠L2PS0−∠L2S0P=180°−75°−50°=55°
∆L2L1S1:∠L2L1S1=∠S0L2S1−∠L2S1L1=55°−25°=30°
∆S0L2S1:sin95°S1L2=sin55°S0S1
S1L2=S0S1sin55°sin95°
∆S1L1L2:sin25°L1L2=sin30°S1L2
L1L2=S1L2sin30°sin25°=S0S1sin55°sin95°⋅sin30°sin25°≈15.42 (miles)
Comments
Dear Raffy, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!
Excellence