Question

Ali sails on a bearing of N20E for 18 km then S85E for 6 km . How far is he from his starting point?

Answer 1

Ali sails on a bearing of N-20-E for 18 km

Then he sails S-85-E for 6 km

Consider a triangle ABC

C = 105 degree

side a = 6 km   

side b = 18 km

Apply cosine rule ------> c^2 = a^2 + b^2 − 2ab cos(C)

-------> c = sqrt [ a^2 + b^2 − 2ab cos(C) ]

-------> c = sqrt [ 6^2 + 18^2 − 2*6*18 cos( 105 ) ]

-------> c = sqrt [ 36 + 324 − 216 cos( 105 ) ]

  -------> c = sqrt [ 360 − 216 cos(105) ]

Now cos( 105 ) = cos( 90 + 15 ) = - sin( 15 ) = - sin( 45 - 30 )

sin( A - B ) = sin A cos B - cos A sin B

- sin( 45 - 30 ) = - [ sin 45 cos 30 - cos 45 sin 30 ]

Thereore , c = sqrt [ 360 − 216 { - [ sin 45 cos 30 - cos 45 sin 30 ] } ]

c = sqrt [ 360 − 216 { - [ ( 1/√2 ) (  √3/2 ) - ( 1/√2 ) ( 1/2 ) ] } ]

  c = sqrt [ 360 + 216 [ ( 1/√2 ) (  √3/2 ) - ( 1/√2 ) ( 1/2 ) ] ]

c = sqrt [ 360 + 216 * 0.258 ]    

c = sqrt [ 415.728 ]

c = 20.389 km

He is about 20.389 km from his starting point.