4MA0 Higher June 2013 Paper 3 Q18

EdexcelHigherOld spec3 marksAdvanced Trigonometry Constructions & Bearings

P due North of Q on a coast; ship S with PS = 2.9 km; angle 62° between north at P and PS, angle 36° between north at Q and QS

Diagram NOT accurately drawn

\(P\) and \(Q\) are two points on a coast.
\(P\) is due North of \(Q\).
A ship is at the point \(S\).
\(PS = 2.9\) km.
The bearing of the ship from \(P\) is 062°
The bearing of the ship from \(Q\) is 036°

Calculate the distance \(QS\).
Give your answer correct to 3 significant figures.

(3)

Mark scheme

Scheme	Marks
\(\dfrac{2.9}{\sin 36^\circ} = \dfrac{QS}{\sin(180 - 62)^\circ}\)	M1
\((QS =)\ \dfrac{2.9 \sin \text{“}118\text{”}^\circ}{\sin 36^\circ}\) oe	M1
4.36	A1
	(3)

Notes

M1: for correct substitution into the Sine Rule

M1: for correct rearrangement (there may be partial evaluation)

A1: for awrt 4.36

Condone use of 62 instead of 118