In the diagram, ABC is a right-angled triangle - OCR - GCSE Maths - Question 13 - 2018 - Paper 5

Since AP = 20 m and AB = AC + PB, we need to find AB first. Using the Pythagorean theorem for right triangle ABC:

$$AB^2 = AC^2 + BC^2$$

Substituting the known values:

$$AB^2 = 20^2 + 40^2 = 400 + 1600 = 2000$$

So,

$$AB = \sqrt{2000} = 20 \sqrt{10}$$

Thus, we can find PB:

$$PB = AB - AP = 20 \sqrt{10} - 20 = 20(\sqrt{10} - 1)$$

To express this in the required form $a (\sqrt{3} - b)$, we can rewrite it as:

$$20 (\sqrt{10} - 1) = 20(\sqrt{3} - 1) \cdot \frac{\sqrt{3}}{\sqrt{3}} = 20(\sqrt{3} \cdot \sqrt{3} - \sqrt{3}) = 20(\sqrt{3} \cdot \sqrt{3} - 3) = 20(\sqrt{3} - 1)$$

Thus, the final answer is:

$a = 20$ and $b = 1$.