Question #139617
From a ship two lighthouses bear N 40o
E. After the ship has sailed 15 miles on a course of 135o
, they bear
10o
and 345o
, respectively. Find the distance between the two lighthouses.
1
Expert's answer
2020-10-22T17:46:32-0400

S0S1=15S_0S_1=15 (miles)

L1S0N0=L2S0N0=40°\angle L_1S_0N_0=\angle L_2S_0N_0=40\degree

L2S0P=50°\angle L_2S_0P=50\degree

N0S0S1=135°\angle N_0S_0S_1=135°

L2S0S1=135°40°=95°\angle L_2S_0S_1=135°-40°=95°

L1S1N1=10°\angle L_1S_1N_1=10°

N1S1L2=360°345°=15°\angle N_1S_1L_2=360°-345°=15°

TPS1:TPS1=90°15°=75°=L2PS0∆TPS_1: \angle TPS_1=90°-15°=75°=\angle L_2PS_0

PL2S0:PL2S0=180°L2PS0L2S0P=180°75°50°=55°∆PL_2S_0:\angle PL_2S_0=180°-\angle L_2PS_0-\angle L_2S_0P=180°-75°-50°=55°

L2L1S1:L2L1S1=S0L2S1L2S1L1=55°25°=30°∆L_2L_1S_1: \angle L_2L_1S_1=\angle S_0L_2S_1-\angle L_2S_1L_1=55°-25°=30°


S0L2S1:S1L2sin95°=S0S1sin55°∆S_0L_2S_1: \frac{S_1L_2}{sin95°}=\frac{S_0S_1}{sin55°}

S1L2=S0S1sin95°sin55°S_1L_2=S_0S_1\frac{sin95°}{sin55°}


S1L1L2:L1L2sin25°=S1L2sin30°∆S_1L_1L_2: \frac{L_1L_2}{sin25°}=\frac{S_1L_2}{sin30°}

L1L2=S1L2sin25°sin30°=S0S1sin95°sin55°sin25°sin30°15.42L_1L_2=S_1L_2\frac{sin25°}{sin30°}=S_0S_1\frac{sin95°}{sin55°}\cdot \frac{sin25°}{sin30°}\approx15.42 (miles)


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Comments

Assignment Expert
23.10.20, 19:51

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Raffy
23.10.20, 01:05

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