Basic Trigonometry: WJEC GCSE Numeracy Higher June 2018 Unit 2 Q11

Numeracy Higher June 2018 Unit 2 Q11

WJECHigherCurrent spec8 marksBasic Trigonometry Pythagoras

11. A sensor can detect any movement up to a distance of 6·5 m.

Diagram not drawn to scale

(a) A storeroom is in the shape of a cuboid, as shown below.
The sensor is placed at A, so that

Diagram not drawn to scale

Will the sensor be able to detect movement at B?
You must show all your working. [5]

(b) Show that \(B\hat{A}E\) = 61·3°, correct to 1 decimal place. [3]

Mark scheme (a)

Scheme	Marks	Comments
(BE² =) 2² + 5·5² BE² = 34·25 or (BE =) √34·25 or (BE =) 5·8(5…)	M1 A1
(AB² =) 34·25 + 3·2² or 5·8(5…)² + 3·2²	M1	FT ‘their derived BE’
(AB =) 6·67(00..) (m)	A1	Accept an answer that would round to 6·67 (m) Accept an answer of 6·7 (m) provided √44·42(...) to √44·55(...) seen in workings Alternative method: M1 for (AD² =) 3·2² + 5·5² A1 for AD² = 40·49 or (AD =) √40·49 or (AD =) 6·3(6…) M1 for (AB² =) 40·49 + 2² or 6·3(6…)² + 2² FT ‘their derived AD’ A1 for (AB =) 6·67(00..) (m) Accept an answer that would round to 6·67 (m) Accept an answer of 6·7 (m) provided √44·42(...) to √44·55(...) seen in workings
(Dist = 6·67(00..) – 0·2 =) 6·47(00..) (m) AND e.g. ‘Yes, it will detect movement at B’	E1	FT correct interpretation of ‘their 6·67(00..)’ – 0.2 provided M1M1 awarded Allow ‘Yes as distance < 6.5 (m)’

Mark scheme (b)

Scheme	Marks	Comments
(BÂE =) cos^-1(3.2 ÷ 6·67(00..) or tan⁻¹ (5·8(523…) ÷ 3·2) or sin^-1 (5·8(523…) ÷ 6·67(00..))	M2	M1 for cos (BÂE) = (3.2 ÷ 6·67(00…) or tan (BÂE) = (5·8(523..) ÷ 3·2) or sin (BÂE) = (5·8(523..) ÷ 6·67(00..)) FT “their derived 5·8(523..) and 'their derived 6·67(00…)’ Do not allow use of 20 for ‘their 6·67(00…)’
(BÂE =) 61·3(306…) (°)	A1	CAO Accept an answer that rounds to 61·3(°) Alternative method: M2 for cos^-1\(\dfrac{6{\cdot}67(...)^{2} + 3{\cdot}2^{2} - 5{\cdot}8(5\ldots)^{2}}{2 \times 6{\cdot}67(...) \times 3{\cdot}2}\) M1 for cos(angle) = \(\dfrac{6{\cdot}67(...)^{2} + 3{\cdot}2^{2} - 5{\cdot}8(5\ldots)^{2}}{2 \times 6{\cdot}67(...) \times 3{\cdot}2}\) FT “their derived 5·8(523..) and 'their derived 6·67(00…)’ Do not allow use of 20 for ‘their 6·67(00…)’ A1 for (BÂE =) 61·3(306…) (°) CAO Accept an answer that rounds to 61·3(°)