Consider the triangle ABC where ${\rm{B\hat AC}} = 70^\circ$, AB = 8 cm and AC = 7 cm. The point D on the side BC is such that $\frac{{{\text{BD}}}}{{{\text{DC}}}} = 2$.

Determine the length of AD.

use of cosine rule: ${\text{BC}} = \sqrt {({8^2} + {7^2} - 2 \times 7 \times 8\cos 70)} = 8.6426 \ldots$     (M1)A1

Note: Accept an expression for ${\text{B}}{{\text{C}}^2}$.

${\text{BD}} = 5.7617 \ldots \,\,\,\,\,{\text{(CD}} = 2.88085 \ldots )$     A1

use of sine rule: $\hat B = \arcsin \left( {\frac{{7\sin 70}}{{{\text{BC}}}}} \right) = 49.561 \ldots ^\circ \,\,\,\,\,(\hat C = 60.4387 \ldots ^\circ )$     (M1)A1

use of cosine rule: ${\text{AD}} = \sqrt {{8^2} + {\text{B}}{{\text{D}}^2} - 2 \times {\text{BD}} \times 8\cos B}  = 6.12{\text{ (cm)}}$     A1

Note: Scale drawing method not acceptable.

[6 marks]

Well done.