assuming they both left from P at the same time ....
you have a clear case of the cosine law.
x^2 = 200^2 + 160^2 - 2(200)(160)cos150°
Careful with the cos150° , it will be negative
A ship P sails at an average speed of 50km/hr on a bearing of N55 degree E from a point R.At the same time another ship Q sails at an average speed of 40km/hr on a bearing of S25 degreeW.How far are they apart after 4hours
2 answers
All angles are measured CW from +Y-axis,
Dp = 50km/hr[55o] * 4hr. = 200km[55o] = Distance of ship P,
Dq = 40km/hr[205o] * 4hr. = 160km[205o] = Distance of ship Q.
D = 200km[55o] - 160km[205o],
X = 200*sin55 - 160*sin205 = 231.4 km,
Y = 200*Cos55 - 160*Cos205 = 259.7 km,
D = Sqrt(X^2+Y^2) = 347.8 km. = Distance between P and Q after 4 hrs.
Tan A = X/Y.
Dp = 50km/hr[55o] * 4hr. = 200km[55o] = Distance of ship P,
Dq = 40km/hr[205o] * 4hr. = 160km[205o] = Distance of ship Q.
D = 200km[55o] - 160km[205o],
X = 200*sin55 - 160*sin205 = 231.4 km,
Y = 200*Cos55 - 160*Cos205 = 259.7 km,
D = Sqrt(X^2+Y^2) = 347.8 km. = Distance between P and Q after 4 hrs.
Tan A = X/Y.
2 answers