4MA0 Higher June 2014 Paper 3R Q18

Diagram NOT accurately drawn

The diagram shows a cuboid \(ABCDEFGH\).
\(AB = 8\) cm, \(AF = 6\) cm and \(FC = 16\) cm.

(a) Find the length of \(BC\).
Give your answer correct to 3 significant figures. (3)

(b) Find the size of the angle between the line \(FC\) and the plane \(ABGF\).
Give your answer correct to 1 decimal place. (2)

Mark scheme (a)

Scheme	Marks
\(6^2 + 8^2 + BC^2 = 16^2\) or \(BC^2 = 16^2 - 6^2 - 8^2\) or \(36 + 64 + BC^2 = 256\) or \(BC^2 = 256 - 36 - 64\) or \(FB^2 = 100\) oe or \(FB = \sqrt{100}\) oe or \(AC^2 = 220\) oe or \(AC = \sqrt{220}\) oe or \(GC^2 = 192\) oe or \(GC = \sqrt{192}\) oe	M1
\(BC = \sqrt{16^2 - 6^2 - 8^2}\) or \(BC = \sqrt{256 - 36 - 64}\) or \(BC = \sqrt{156}\) or \(BC = \sqrt{16^2 - 10^2}\) or \(BC = \sqrt{256 - 100}\) or \(BC = \sqrt{220 - 8^2}\) oe or \(BC = \sqrt{192 - 6^2}\) oe	M1
12.4899… Answer: 12.5	A1
	(3)

M1: for use of Pythagoras in 3D shape or a correct value for \(FB\) or \(FB^2\) or AC or AC²

M1: For a correct method to find BC.

(Condone \(2\sqrt{55}^2\) rather than \((2\sqrt{55})^2\) for 220 for method mark)

A1: for answer which rounds to 12.5

Mark scheme (b)

Scheme	Marks
\(\cos BFC = \dfrac{10}{16}\) or \(\sin BFC = \dfrac{\text{“}12.5\text{”}}{16}\) or \(\tan BFC = \dfrac{\text{“}12.5\text{”}}{10}\)	M1
51.3	A1
	(2)
	(5 marks)

M1: A correct equation for the correct angle.

A1: for awrt 51.3 or 51.4