The buoy is situated at B, 43 m from the pier P and 26 m from point Q, as shown in the diagram above - Leaving Cert Mathematics - Question c - 2021

Substituting the values gives:

$60^2 = 26^2 + 43^2 - 2(26)(43)\cos(A)$

Calculating the squares:

$3600 = 676 + 1849 - 2(26)(43)\cos(A)$

Now simplifying:

$3600 = 2525 - 2234\cos(A)$

This leads to:

$2234\cos(A) = 2525 - 3600$ $2234\cos(A) = -1075$

Next, we can solve for cos(A):

$\cos(A) = \frac{-1075}{2234}$

Calculating this gives:

$\cos(A) \approx -0.481$

Now, using the inverse cosine function to find angle A:

$A = \cos^{-1}(-0.481)$

Calculating this value gives:

$A \approx 118.74°$

The final answer is:

$\angle(QBP) \approx 118.74°$

This is correct to two decimal places.