A boy leaves station X on a bearing of 035° to station y which is 21km away. He then travels to another…

Analyze the question

The boy goes from X to Y (21 km, bearing 035°), then from Y to Z on bearing 125°. Z is due East of X. We need XZ.

Set up the geometry

• At X, the bearing of Y is 035°, so the angle between North and XY is 35°. Z is due East (090°), so angle YXZ = 90° − 35° = 55°.
• At Y, the bearing of Z is 125°. The back-bearing from Y to X is 035° + 180° = 215°. So the angle XYZ = 215° − 125° = 90°.
• Angle at Z = 180° − 55° − 90° = 35°.

So triangle XYZ is right-angled at Y.

Solve

XY is adjacent to the 55° angle at X, and XZ is the hypotenuse:

cos 55° = XY / XZ

XZ = 21 / cos 55° = 21 / 0.5736 = 36.61 km

Check: Using the 35° angle at Z, sin 35° = XY / XZ, so XZ = 21 / 0.5736 = 36.61 km. Both methods agree.

**Answer: E (36.

Tip: For bearing questions, draw a North line at every station. Use parallel North lines (alternate and co-interior angles) to find the interior angles, and always compute the back-bearing by adding or subtracting 180°. Here, noticing the 90° angle at Y made it a simple right-triangle problem.