The diagram shows a semi-circle BAC, with diameter BC = 34 cm. AD and BCD are straight lines. Given that $\cos\angle ABC = \dfrac{8}{17}$, without solving for angle ABC and angle ACD, find the exact value of $\cos\angle ACD$. Answer ...

1. Identify the right angle in triangle ABC and interpret the given cosine ratio.

Since $BC$ is the diameter of the semicircle, $\angle BAC = 90^\circ$.

In right-angled triangle $ABC$,
$\cos\angle ABC=\frac{AB}{BC}=\frac{8}{17}.$
Given $BC=34$ cm,
$\frac{AB}{34}=\frac{8}{17} \quad\Rightarrow\quad AB=16\text{ cm}.$

2. Find the length of AC using Pythagoras' theorem.

Using Pythagoras' theorem in $\triangle ABC$,
$AC^2=BC^2-AB^2$
$AC^2=34^2-16^2=1156-256=900$
$AC=30\text{ cm}.$

3. Find the cosine of the interior angle ACB.

In right-angled triangle $ABC$,
$\cos\angle ACB=\frac{AC}{BC}=\frac{30}{34}=\frac{15}{17}.$

4. Use the straight line BCD to relate angle ACD to angle ACB.

Since $B$, $C$ and $D$ are collinear, $\angle ACB$ and $\angle ACD$ are supplementary.

Hence
$\cos\angle ACD=-\cos\angle ACB=-\frac{15}{17}.$