Please help me with the following mathematical question

Question

There is a rescue boat that is already out at sea at R. However, it receives distress calls from the yacht and another boat, at the same time.
-The rescue boat must go to the nearest distress call first.
-The boat is located at B, 350 km away, on a bearing of 120°T from R.
-The yacht is located at Y on a bearing of 210°T and 170 km away from B.

 

 

Which is the nearest distress call?

a) Show that ∠RBY is a right-angle.
i. The North/South lines at R and B are parallel lines. Use your knowledge of co-interior angles to find: ∠PBR = ___°

ii. Using the number of degrees in a revolution, show ∠RBY = 90°

 

 

Answer 1

a) This answer requires a little drawing.  By general definition of bearing, 0° is due north and moves in a clockwise direction.  Draw x and y vertices where R is the origin.  From this we get that the boat is 30° S from East at 350 km.  Draw a vertical line on the graph that goes through point B and call it l.  We can see that we've created a right triangle that includes points B and R but includes the line for East (our x-axis).  This is a 30, 60 right triangle.  By looking at Y, 210° is 30° West from S.  Since line l is 180° and we have just calculated 30° and the angle from the earlier triange is 60° we get that ∠RBY is 90°.

 

To finish the problem since the distance from R to Y is the hypotenuse, it has to be the longest distance on the triangle.