Problem Statement
A ship departs harbor $H$ and sails $18\text{ km}$ on a bearing of $040^\circ$ to point $A$. The ship then changes course and sails $25\text{ km}$ on a bearing of $130^\circ$ to point $B$.
**(a)** [3 marks] Show that angle $HAB$ is $90^\circ$.
**(b)** [3 marks] Calculate the direct distance from harbor $H$ to point $B$, and find the bearing of $B$ from $H$.
Verified Solution & Marking Scheme
Part (a) [M1]
Angle of line $HA$ from South at point $A$ is $180^\circ - 40^\circ = 140^\circ$ (or alternate interior angles $40^\circ$ from south).
Part (a) [M1]
The bearing of $B$ from $A$ is $130^\circ$. The angle between South and bearing $130^\circ$ is $180^\circ - 130^\circ = 50^\circ$.
Part (a) [A1]
Angle $HAB = 40^\circ + 50^\circ = 90^\circ$ (AG)
Part (b) [M1]
Using Pythagoras on right triangle $HAB$: $HB = \sqrt{18^2 + 25^2} = \sqrt{324 + 625} = \sqrt{949} \approx 30.8\text{ km}$
Part (b) [M1]
Angle $\angle AHB = \arctan\left(\frac{25}{18}\right) = 54.2^\circ$
Part (b) [A1]
Bearing of $B$ from $H$ is $040^\circ + 54.2^\circ = 094.2^\circ$ (or $094^\circ$ to 3 s.f.)