1.	R =   10√3/2 i – 5j	M1 A1
	Using P = 7j and Q = R – P to obtain Q = 5√3i – 12j	M1 A1
	Magnitude = √[(5√3)^2 + 12^2] ≈ 14.8 N (AWRT)	M1 A1
	angle with i =   arctan (12/5√3) ≈ 64.2°	M1 A1
	bearing ≈ 144° (AWRT)	A1	9
 
	Alternative method

	Vector triangle correct	B1
	Q^2 = 10^2 + 7^2 + 2 × 10 × 7 cos 60	M1 A1
	Q  ≈ 14.8 N (AWRT)	A1
		M1 A1
	⇒ θ = 35.8,   ⇒ bearing 144 (AWRT)	M1 A1, A1
[9]

El Alsson British International School