Answer to Question #139617 in Trigonometry for Raffy

Question

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.

Answer 1

$S_0S_1=15$ (miles)

$\angle L_1S_0N_0=\angle L_2S_0N_0=40\degree$

$\angle L_2S_0P=50\degree$

$\angle N_0S_0S_1=135°$

$\angle L_2S_0S_1=135°-40°=95°$

$\angle L_1S_1N_1=10°$

$\angle N_1S_1L_2=360°-345°=15°$

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

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

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

$∆S_0L_2S_1: \frac{S_1L_2}{sin95°}=\frac{S_0S_1}{sin55°}$

$S_1L_2=S_0S_1\frac{sin95°}{sin55°}$

$∆S_1L_1L_2: \frac{L_1L_2}{sin25°}=\frac{S_1L_2}{sin30°}$

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