The diagram shows the triangles $BCD$ and $ABD$, with some measurements given - Leaving Cert Mathematics - Question 5 - 2015

To find the area $A$ of triangle $BCD$, we can use the formula:

$A = \frac{1}{2} \cdot a \cdot b \cdot \sin(c)$

Where:

• $a = 10$ m
• $b = |BC| \approx 11.39$ m
• $c = \angle BDC = 28^\circ = 180^\circ - (42^\circ + 110^\circ)$

Thus:

$A = \frac{1}{2} \cdot 10 \cdot 11.39 \cdot \sin(28^\circ)$

Calculating:

1. $\sin(28^\circ) \approx 0.4695$
2. Substitute into the equation:

$A \approx \frac{1}{2} \cdot 10 \cdot 11.39 \cdot 0.4695 \approx 42.78 \text{ m}^2$

Therefore, the area of triangle $BCD$ is approximately $42.78$ m².

To find $|AB|$, we will also use the cosine rule. The triangle $ABD$ implies:

$|AB|^2 = |AD|^2 + |BD|^2 - 2 |AD| |BD| \cos(75^\circ)$

Given that:

• $|AD| = 10$ m
• $|BD| = 16$ m
• $75^\circ = 180^\circ - (63^\circ + 42^\circ)$

Thus:

1. Plugging into the cosine rule:

$|AB|^2 = 10^2 + 16^2 - 2(10)(16)\cos(75^\circ)$

2. With $\cos(75^\circ) \approx 0.2588$, we have:

$|AB|^2 = 100 + 256 - 320(0.2588)$

3. Calculating further:

$|AB|^2 = 356 - 82.816 = 273.178$ 4. Finally, taking the square root:

$|AB| \approx \sqrt{273.178} \approx 16.53 \text{ m}$

Therefore, $|AB| = 16.53$ m.