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.

${\text{A}}{{\text{C}}^2} = {7.8^2} + {10.4^2}$    (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]

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