Higher June 2023 Paper 2R Q22

EdexcelHigherCurrent spec5 marksAdvanced Trigonometry Pythagoras

22 Here is a cuboid \(ABCDEFGH\)

Diagram NOT accurately drawn

\(AB = 15\) cm \(\qquad BC = 4\) cm \(\qquad CF = 9\) cm

(a) Work out the length of \(BE\)
Give your answer correct to 3 significant figures. (2)

Here is a cuboid \(PQRSTUVW\)

Diagram NOT accurately drawn

\(PR = 42\) cm

The size of the angle between \(PU\) and the plane \(PQRS\) is \(30^\circ\)

\(M\) is the midpoint of \(PR\)

(b) Work out the size of angle \(UMR\)
Give your answer correct to 3 significant figures. (3)

Mark scheme (a)

Scheme	Marks
\(\sqrt{4^2 + 9^2 + 15^2}\;\left(= \sqrt{322} = 17.9(443\ldots)\right)\) or \(\sqrt{15^2 + 4^2}\;\left(= \sqrt{241} = 15.5(241\ldots)\right)\) and \(\sqrt{9^2 + \left(\text{``}\sqrt{241}\text{''}\right)^2}\;\left(= \sqrt{322} = 17.9(443\ldots)\right)\)	M1
17.9	A1
	(2)

A1: awrt 17.9

Mark scheme (b)

Scheme	Marks
\((UR =)\;42\tan 30\) (= \(14\sqrt{3}\) = 24.2(487…)) or \((UR =)\;\dfrac{42 \times \sin 30}{\sin(90 - 30)}\) (= \(14\sqrt{3}\) = 24.2(487…))	M1
\(\tan(UMR) = \left(\dfrac{\text{``}24.248\ldots\text{''}}{42 \div 2}\right)\) or \(\tan(UMR) = \left(\dfrac{\text{``}24.248\ldots\text{''}}{21}\right)\) or \(\tan(UMR) = \left(\dfrac{\text{``}14\sqrt{3}\text{''}}{21}\right)\) or \((UM =)\;\sqrt{\left(\dfrac{42}{2}\right)^2 + \left(\text{``}14\sqrt{3}\text{''}\right)^2}\;\left(= 7\sqrt{21} = 32.0(780\ldots)\right)\) and \(\sin(UMR) = \left(\dfrac{\text{``}14\sqrt{3}\text{''}}{\text{``}7\sqrt{21}\text{''}}\right)\) or \(\cos(UMR) = \left(\dfrac{21}{\text{``}7\sqrt{21}\text{''}}\right)\)	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 49.1	A1
	(3)
	(5 marks)

A1: awrt 49.1