IBDP Maths AA: Topic SL 3.3: Applications of right and non-right angled trigonometry, including Pythagoras’s theorem: IB style Questions HL Paper 1

Question

ABCD is a quadrilateral where \({\text{AB}} = 6.5,{\text{ BC}} = 9.1,{\text{ CD}} = 10.4,{\text{ DA}} = 7.8\) and \({\rm{C\hat DA}} = 90^\circ \). Find \({\rm{A\hat BC}}\), giving your answer correct to the nearest degree.

▶️Answer/Explanation

Markscheme

\({\text{A}}{{\text{C}}^2} = {7.8^2} + {10.4^2}\)    (M1)

\({\text{AC}} = 13\)    (A1)

use of cosine rule eg, \(\cos ({\rm{A\hat BC}}) = \frac{{{{6.5}^2} + {{9.1}^2} – {{13}^2}}}{{2(6.5)(9.1)}}\)     M1

\({\rm{A\hat BC}} = 111.804 \ldots ^\circ {\text{ }}( = 1.95134 \ldots )\)    (A1)

\( = 112^\circ \)    A1

[5 marks]

Examiners report

Well done by most candidates. A small number of candidates did not express the required angle correct to the nearest degree.

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