The diagram represents a field, $ABCD$. In it, $AD = 180\,\text{m}$ and $AC = 240\,\text{m}$. In addition, angle $ABC = 50^{\circ}$ and angle $ACB = 85^{\circ}$.
(a)[3]
Calculate $AB$ by using the sine rule.
(b)[3]
Show that angle $CAD = 33.75^{\circ}$, correct to 2 decimal places.
(c)[5]
Calculate the length of $BD$.
(d(i))[1]
Determine the bearing of $B$ from $A$.
(d(ii))[2]
Determine the reverse bearing, that is, the bearing of $A$ from $B$.
Worked solution & mark scheme
This 14-mark question has a full step-by-step worked solution and mark scheme. One marking point: “The required expression is $\frac{240\sin85}{\sin50}$.” …