Advanced Trigonometry: WJEC GCSE Numeracy Higher November 2020 Unit 2 Q11

Numeracy Higher November 2020 Unit 2 Q11

WJECHigherCurrent spec8 marksAdvanced Trigonometry Circles & Sectors

11. A square trapdoor has sides of length 110 cm.
When the trapdoor is fully opened, it makes an angle of 85º with the horizontal.

Square trapdoor of side 110 cm opened at 85° to the horizontal, with a dashed arc showing the path of corner A

Diagram not drawn to scale

(a) Calculate the distance that point A travels as the trapdoor moves from being closed to being fully opened. [2]

(b) A thin rod is used to hold the trapdoor open in this position.
The rod goes from the closed position of point A to a point on the edge of the trapdoor, 16 cm from point A.

The open trapdoor at 85° with a rod from the closed position of A to a point on the edge 16 cm from A

Diagram not drawn to scale

Calculate both the length of the rod and the angle the rod makes with the horizontal. [6]

Length of rod = ………………… cm
Angle the rod makes with the horizontal = ………………… º

From the formula list: In any triangle \(ABC\): Sine rule \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\); Cosine rule \(a^2 = b^2 + c^2 - 2bc\cos A\)

Scheme	Marks	Comments
(Distance travelled =) \(\dfrac{85}{360} \times 2 \times \boldsymbol{\pi} \times 110\)	M1
= 163 to 163.21 (cm)	A1	Or \(\dfrac{935\boldsymbol{\pi}}{18}\) or equivalent

Mark scheme (b)

Scheme	Marks	Comments
(Length of rod =) √(110² + (110 – 16)² − 2×110×(110 – 16)×cos85(º)) (=√19133.6(192…))	M2	M1 for length² = 110²+(110–16)² −2×110×(110–16)×cos85(º)
= 138(.324…) (cm)	A1
(Angle =) \(\sin^{-1}\!\left(\dfrac{\sin 85(^\circ)}{138(.324\ldots)} \times (110 - 16)\right)\) OR (Angle =) \(\cos^{-1}\!\left(\dfrac{110^{2} + 138(.324\ldots)^{2} - (110 - 16)^{2}}{2 \times 110 \times 138(.324\ldots)}\right)\) OR full complete method using the sine rule to find the angle at the top of the triangle and then use of angles in a triangle	M2	FT ‘their derived 138(.324…)’ M1 for \(\dfrac{\sin(\text{angle})}{(110 - 16)} = \dfrac{\sin 85(^\circ)}{138(.324\ldots)}\) or equivalent or sin (angle) = \(\dfrac{\sin 85(^\circ)}{138(.324\ldots)}\) × (110 – 16) OR M1 for (110−16)² =110²+138(.324…)² –2×110×138(.324…)×cos (angle) or cos (angle) = \(\dfrac{110^{2} + 138(.324\ldots)^{2} - (110-16)^{2}}{2 \times 110 \times 138(.324\ldots)}\) OR M1 for finding the angle at the top of the triangle (52·2(6…) to 53 (º))
= 42·6 to 42·7(3…) (º)	A1	Accept 43 (º) from correct working