Question

.Suppose I am in a boat and I travel at the bearing N70E at 30 knots for 4 hours. Then, I turn 90 degrees clockwise and travel for 5 hours at the same speed. Find my bearing relative to the dock.

Answer 1

Consider a boat is travelling at  N70E at 30 knot for 4 hours

So 1 knot = 1.85 kmph

30 knot = 30*1.85 kmph = 55.5 kmph

So, in 1 hour boat travels 55.5 km

in 4 hours boat travels 55.5*4 = 222 km

Then it turs 90 degree clockwise and travels for 5 hours at same speed ---> 30 knot

So,  

30 knot = 30*1.85 kmph = 55.5 kmph

So, in 1 hour boat travels 55.5 km

in 5 hours boat travels 55.5*5 = 277.5 km

Consider right triangle ABC

Apply Pythagoras Property

b = sqrt [ a^2 + c^2 ]

b = sqrt [ 227.5^2 + 222^2 ] = sqrt [ 51756.25 + 49284 ]

= sqrt [ 101040.25 ] = 317.86 km

Now Apply sine rule

sin( A ) / a = sin( B ) / b

=> sin( A ) / 227.5 = 1 / 317.86 ........ sin( 90 ) = 1

=> sin( A ) = 227.5 / 317.86 = 0.7157

=> A = arcsin( 0.7157 ) = 45.7 degree

So relative bearing angle is 180 - 70 - 45.7 = 64.3 degree

So relative bearing to the dock is ------> S - 64.3 - E