4400 Higher November 2006 Paper 4 Q14

EdexcelHigherOld spec7 marksBasic Trigonometry

Triangle LMN right-angled at N with MN = 5.4 cm horizontal and LN = 9.3 cm vertical

Diagram NOT accurately drawn

Triangle \(LMN\) is right-angled at \(N\).
\(MN = 5.4\) cm and \(LN = 9.3\) cm.

(a) Work out the size of angle \(LMN\).
Give your answer correct to 1 decimal place. (3)

The length of \(MN\) is 5.4 cm, correct to 2 significant figures.

(b)

(i) Write down the upper bound of the length of \(MN\).

(ii) Write down the lower bound of the length of \(MN\). (2)

The length, 5.4 cm, of \(MN\) and the length, 9.3 cm, of \(LN\), are each correct to 2 significant figures.
The line \(MN\) is horizontal and the line \(LN\) is vertical.

Mark scheme (a)

Scheme	Marks
\(\tan \angle LMN = \dfrac{9.3}{5.4}\)	M1
\(\frac{9.3}{5.4}\) or 1.722...	A1
59.9	A1
	(3)

Notes

M1: \(\sin LMN = \dfrac{9.3}{\sqrt{9.3^2 + 5.4^2}}\) or cos etc M1A1

A1: for 59.85 − 59.9

Scheme	Marks
(i) Answer: 5.45	B1
(ii) Answer: 5.35	B1
	(2)

Scheme	Marks
\(\dfrac{9.35}{\text{“}5.35\text{”}}\)	M1
1.74766...	A1
	(2)
	(7 marks)