ABC is a right-angled triangle with $\angle ABC = 90^\circ$, $AB = 8$ cm and $BC = 15$ cm. Find the value of $\cos \angle BCD$. Answer $\cos \angle BCD =$ ___

1. Find the length of the hypotenuse $AC$.

Since $\angle ABC = 90^\circ$,
$AC^2 = AB^2 + BC^2$
$AC^2 = 8^2 + 15^2 = 64 + 225 = 289$
$AC = 17\text{ cm}$

2. Relate the exterior angle to the interior angle and find its cosine.

For $\angle BCA$,
$\cos \angle BCA = \frac{BC}{AC} = \frac{15}{17}.$
Since $CD$ is the extension of $AC$,
$\angle BCD = 180^\circ - \angle BCA.$
Hence
$\cos \angle BCD = -\cos \angle BCA = -\frac{15}{17}.$