The triangle ABC is equilateral of side 3 cm. The point D lies on [BC] such that BD = 1 cm.

Find \(\cos {\rm{D\hat AC}}\).

METHOD 1

\({\text{A}}{{\text{D}}^2} = {2^2} + {3^2} - 2 \times 2 \times 3 \times \cos 60^\circ \)     M1

(or \({\text{A}}{{\text{D}}^2} = {1^2} + {3^2} - 2 \times 1 \times 3 \times \cos 60^\circ \))

Note:     M1 for use of cosine rule with 60° angle.

\({\text{A}}{{\text{D}}^2} = 7\)     A1

\(\cos {\rm{D\hat AC}} = \frac{{9 + 7 - 4}}{{2 \times 3 \times \sqrt 7 }}\)     M1A1

Note:     M1 for use of cosine rule involving \({\rm{D\hat AC}}\).

\( = \frac{2}{{\sqrt 7 }}\)     A1

METHOD 2

let point E be the foot of the perpendicular from D to AC

EC = 1 (by similar triangles, or triangle properties)     M1A1

(or AE = 2)

\({\text{DE}} = \sqrt 3 \) and \({\text{AD}} = \sqrt 7 \)   (by Pythagoras)     (M1)A1

\(\cos {\rm{D\hat AC}} = \frac{2}{{\sqrt 7 }}\)     A1

Note:     If first M1 not awarded but remainder of the question is correct award M0A0M1A1A1.

[5 marks]