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]