Two swimmers A and B stand at the same point X, on one shore of a long, still rectangular shaped lake that is 100 m wide, as shown below - Leaving Cert Mathematics - Question 5 - 2020

For Swimmer B, we can also use the sine function:

$$ext{sin}( heta) = \frac{100}{200}$$

So,

$$heta = ext{sin}^{-1}(0.5) = 30°$$

Thus, the angle θ is 30°.

First, let's calculate the horizontal distances for both swimmers:

For Swimmer A:

$$X_A = 141.4 \times ext{cos}(45°) \approx 141.4 \times 0.7071 \approx 100 ext{ m}$$

For Swimmer B:

$$X_B = 155.6 \times ext{cos}(40°) \approx 155.6 \times 0.7660 \approx 119.1 ext{ m}$$

Now, to find the distance d between both swimmers:

$$|X| = |X_A - X_B| = |100 - 119.1| = 19.1 ext{ m} ext{ (approximately 19 m)}$$

Thus, distance d is approximately 19 m.