The bearing of C from A is N 30° E and from B, 50 metres east of A, is N 60° W. The departure of C from A is

ANS:D - 25 m

When we draw triangle ABC, we found angle A,B,C 60,30,90 and AB=50m.

By applying geometry AC = ABcos60---> (1)
And departure of AC will be ACcos60 ---> (2).

From (1) put value of AC in (2) we got;
Departure of AC= (AB cos60) cos60.

50 x 0.5 x 0.5 = 12.5m. B is 50 m east of A, here departure of C with respect to A has to be found hence RB is 30, so sin 30 is correct, but L is not 50. L has to be distance AC, which is from calculation we get it as 25m. I think the answer is 12.5m.

The given distance 50m is for line AB , so you cannot use it directly to solve for departure on line CA.

So using the triangle ABC , given the length of AB=50m and the angles can be computed using the given bearings. The side AC can be computed and AC=25m.

Then, the departure of line AC is 25sin(30)= 12.5m.