Distance Between Ships One ship leaves a port at 1:00 P.M. traveling at 1[a] mph at a heading of [c]5o. At 2:00 P.M. another ship leaves the port traveling at 1[c] mph at a heading of 14[e]o. Find the distance, to the nearest mile, between the ships at 3:00 P.M. Be sure to include a sketch. Note: if [a]=3, then 1[a]=13 mph, if [c]=8, then [c]5o=85o, and if[e]=2, then 14[e]o=142o My PIN : 3 5 8 9 2 [a][b][c][d][e] Work this custom problem about oblique triangles using the techniques you learned in this module. For Ship I Speed=13mph Therefore, distance covered in 2 hours (1 pm to 3 pm)=13*2=26 mile. For Ship I Speed=18mph Therefore, distance covered in 1 hours (2 pm to 3 pm) =13*2=26 mile. Since, angle made by ship is 142 degree and ship I is 85 degree. Therefore, angle A=142-85=57 degree. We have to find a. By cosine formula a 2 b 2 c 2 2bc cos A a 2 182 26 2 2(18)(26) cos 57 0 a 2 324 676 936 cos 57 0 a 2 1000 509.7821 a 2 490.2179 a 22.14 a 22mile( approx ) Therefore, distance between Ships at 3:00 Pm is 22 miles approximately. For Ship I Speed=13mph Therefore, distance covered in 2 hours (1 pm to 3 pm)=13*2=26 mile. For Ship I Speed=18mph Therefore, distance covered in 1 hours (2 pm to 3 pm) =13*2=26 mile. Since, angle made by ship is 142 degree and ship I is 85 degree. Therefore, angle A=142-85=57 degree. We have to find a. By cosine formula a 2 b 2 c 2 2bc cos A a 2 182 26 2 2(18)(26) cos 57 0 a 2 324 676 936 cos 57 0 a 2 1000 509.7821 a 2 490.2179 a 22.14 a 22mile( approx ) Therefore, distance between Ships at 3:00 Pm is 22 miles approximately