Advanced Trigonometry: WJEC GCSE Higher June 2023 Unit 2 Q16

Higher June 2023 Unit 2 Q16

WJECHigherCurrent spec6 marksAdvanced Trigonometry

16. In triangle ABC shown below, AB = 13 cm and BC = 11 cm.
D is a point on AC where BD = 7 cm and DC = 5 cm.

Diagram not drawn to scale

Calculate the size of \(B\hat{A}D\).
You must show all your working. [6]

From the formula list: In any triangle \(ABC\): Cosine rule \(a^2 = b^2 + c^2 - 2bc\cos A\)

Mark scheme

Scheme	Marks	Comments
Initially calculating angle BDC cos BDC = \(\dfrac{5^2 + 7^2 - 11^2}{2 \times 5 \times 7}\) (= –47/70)	M2	M1 for 11² = 5² + 7² – 2×5×7×cos BDC
(BDC =) 132(·17…°)	A1	The correct angles given in: Radians (2·3038… to 2·3073…) or Gradians (146·6666… to 146·8888...). Allow a trial and improvement method for M2A1 but length of side BC must be in the range 10·95(cm) to 11·05(cm). Sight of 28(·13…°) OR 132(·17…°) OR 19(·68…°), calculated from one use of the rearranged cosine rule in triangle BCD and attributed to the wrong angle, stated or seen on the diagram, is M0 SC1.
sin BAD = \(\dfrac{7 \times \sin\,[180 - 132(\cdot 17\ldots^\circ)]}{13}\)	M2	FT ‘their stated or derived BDC’. M1 for \(\dfrac{\sin \text{BAD}}{7} = \dfrac{\sin\,[180 - 132(\cdot 17\ldots^\circ)]}{13}\) o.e.
(BAD = ) Answers in the range 23·4(°) to 23·6(°)	A1	Mark final answer for angle BAD. Allow a correctly rounded answer to a whole number provided an answer in the range 23·4(°) to 23·6(°) is seen.

Scheme	Marks	Comments
Initially calculating angle BCD cos BCD = \(\dfrac{5^2 + 11^2 - 7^2}{2 \times 5 \times 11}\) (=97/110)	M2	M1 for 7² = 5² + 11² – 2×5×11×cos BCD
(BCD =) 28(·13…°)	A1	The correct angles given in: Radians (0·4886… to 0·4911…) or Gradians (31·1111… to 31·2666...). Allow a trial and improvement method for M2A1 but length of side BD must be in the range 6·95(cm) to 7·05(cm). Sight of 28(·13…°) OR 132(·17…°) OR 19(·68…°), calculated from one use of the rearranged cosine rule in triangle BCD and attributed to the wrong angle, stated or seen on the diagram, is M0 SC1.
sin BAD = \(\dfrac{11 \times \sin 28(\cdot 13\ldots^\circ)}{13}\)	M2	FT ‘their stated or derived BCD’. M1 for \(\dfrac{\sin \text{BAD}}{11} = \dfrac{\sin 28(\cdot 13\ldots^\circ)}{13}\) o.e.
(BAD = ) Answers in the range 23·4(°) to 23·6(°)	A1	Mark final answer for angle BAD. Allow a correctly rounded answer to a whole number provided an answer in the range 23·4(°) to 23·6(°) is seen.

Scheme	Marks	Comments
Multistep method to calculate angle BDC (First BCD, then CBD or vice versa) cos BCD = \(\dfrac{5^2 + 11^2 - 7^2}{2 \times 5 \times 11}\) (=97/110) OR cos CBD = \(\dfrac{7^2 + 11^2 - 5^2}{2 \times 7 \times 11}\) (=145/154)	M2	A complete, correct method must be seen for the first M2. M1 for one of the following: 7² = 5² + 11² – 2×5×11×cos BCD OR 5² = 7² + 11² – 2×7×11×cos CBD
(BCD =) 28(·13…°) OR (CBD =) 19(·68…°) sin CBD = \(\dfrac{5 \times \sin\,[28(\cdot 13\ldots^\circ)]}{7}\) OR sin BCD = \(\dfrac{7 \times \sin 19(\cdot 68\ldots^\circ)}{5}\) (CBD =) 19(·68…°) OR (BCD =) 28(·13…°) (BDC = ) 180 – [28(·13…°)+19(·68…°)] = 132(·17…°)	A1	The correct angles given in: Radians (BCD ≈ 0·49 or CBD ≈ 0·34) or Gradians (BCD ≈31·2 or CBD ≈ 21.9).
sin BAD = \(\dfrac{7 \times \sin\,[180 - 132(\cdot 17\ldots^\circ)]}{13}\)	M2	This M2 (or M1 below) is for a complete multistep method, i.e. it includes all the previous steps. FT ‘their stated or derived BDC’. M1 for \(\dfrac{\sin \text{BAD}}{7} = \dfrac{\sin\,[180 - 132(\cdot 17\ldots^\circ)]}{13}\) o.e.
(BAD = ) Answers in the range 23·4(°) to 23·6(°)	A1	Mark final answer for angle BAD. Allow a correctly rounded answer to a whole number provided an answer in the range 23·4(°) to 23·6(°) is seen.

Scheme	Marks	Comments
Multistep method to calculate angle BDC (First BCD, then BDC OR first CBD, then BDC)		Ambiguous case
cos BCD = \(\dfrac{5^2 + 11^2 - 7^2}{2 \times 5 \times 11}\) (=97/110) OR cos CBD = \(\dfrac{7^2 + 11^2 - 5^2}{2 \times 7 \times 11}\) (=145/154)	M2	A complete, correct method must be seen for the first M2. M1 for one of the following: 7² = 5² + 11² – 2×5×11×cos BCD OR 5² = 7² + 11² – 2×7×11×cos CBD
(BCD =) 28(·13…°) OR (CBD =) 19(·68…°) sin BDC = \(\dfrac{11 \times \sin\,[28(\cdot 13\ldots^\circ)]}{7}\) (=47·82…°) OR sin BDC = \(\dfrac{11 \times \sin 19(\cdot 68\ldots^\circ)}{5}\) (=47·82…°) (BDC = 180–47(·82…°) ) = 132(·17…°)	A1	The correct angles given in: Radians (BCD ≈ 0·49 or CBD ≈ 0·34) Gradians (BCD ≈31·2 or CBD ≈ 21.9).
sin BAD = \(\dfrac{7 \times \sin\,[180 - 132(\cdot 17\ldots^\circ)]}{13}\)	M2	This M2 (or M1 below) is for a complete multistep method, i.e. it includes all the previous steps. FT ‘their stated or derived BDC’. M1 for \(\dfrac{\sin \text{BAD}}{7} = \dfrac{\sin\,[180 - 132(\cdot 17\ldots^\circ)]}{13}\) o.e.
(BAD = ) Answers in the range 23·4(°) to 23·6(°)	A1	Mark final answer for angle BAD. Allow a correctly rounded answer to a whole number provided an answer in the range 23·4(°) to 23·6(°) is seen.