a)
NOT TO SCALE
Figure 3
Figure 3 shows the acute-angled triangle $\mathrm{ABC}$ with $\mathrm{AB}=9$ metres, $\mathrm{AC}=10$ metres and $\mathrm{BC}=\sqrt{93}$ metres.
i) Find $\cos \mathrm{A}$ in the form $\frac{p}{q}$ where $p$ and $q$ are integers.
[3]

Step 1 ：Given an acute-angled triangle ABC with AB=9 metres, AC=10 metres and BC=\(\sqrt{93}\) metres, we are asked to find \(\cos A\).

Step 2 ：We can use the law of cosines to find the cosine of angle A. The law of cosines states that for any triangle with sides of lengths a, b, and c, and an angle A opposite side a, the following equation holds: \[a^2 = b^2 + c^2 - 2bc \cos A\]

Step 3 ：In this case, a = BC = \(\sqrt{93}\), b = AB = 9, and c = AC = 10. We can rearrange the equation to solve for \(\cos A\): \[\cos A = \frac{b^2 + c^2 - a^2}{2bc}\]

Step 4 ：Substituting the given values into the equation, we get \(\cos A = \frac{9^2 + 10^2 - (\sqrt{93})^2}{2 \times 9 \times 10}\)

Step 5 ：Solving the above equation, we find that \(\cos A\) is approximately 0.4888888888888889.

Step 6 ：Final Answer: The cosine of angle A is approximately \(\boxed{\frac{49}{100}}\).