In triangle ABC the median through A meets BC at M. Given that AB = 8 cm, AM = 5 cm and AC = 6 cm, use the cosine rule in triangles ABM and AMC to find BC. Also find angle ABC.
I have used the cosine rule to establish :
$BM^2 = MC^2$ So that:
$89 – 80 \cos \angle BAM = 61 – 60 \cos \angle CAM$ but I can proceed no further.
Best Answer
The cosine law should be applied to $\angle AMB $ and $\angle AMC$. For $BM = MC = x$, we have
$$cos \angle AMB = \frac{x^2 + 25 - 64}{10x}$$ $$cos \angle AMC = cos (\pi - \angle AMB) = - cos \angle AMB = \frac{25 + x^2 - 36}{10x}$$
Hence we have $$\frac{x^2 - 39}{10x} = -\frac{x^2 - 11}{10x}$$ $$\implies x = 5$$