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

To find the area of triangle BCD, we first need to calculate angle (\angle DBC):

$\angle DBC = 180^\circ - (42^\circ + 110^\circ) = 28^\circ$

Now applying the area formula for a triangle:

$A = \frac{1}{2} a b \sin C$

Where:

• a = 16 m
• b = |BC| \approx 11.39 m
• C = \angle DBC = 28^\circ

Calculating the area:

$A = \frac{1}{2} \cdot 16 \cdot 11.39 \cdot \sin(28^\circ)$

After calculating:

• (\sin(28^\circ) \approx 0.4695)

Substituting gives:

$A \approx \frac{1}{2} \cdot 16 \cdot 11.39 \cdot 0.4695 \approx 42.78 m^2$

Thus, the area of triangle BCD is 42.78 m² (to two decimal places).

To find | AB |, we use the Law of Cosines:

$|AB|^2 = a^2 + b^2 - 2ab \cos A$

Where:

• a = 10 m
• b = 16 m
• A = \angle BDA = 63^\circ + 42^\circ = 105^\circ

Substituting in the values:

$|AB|^2 = 10^2 + 16^2 - 2 \cdot 10 \cdot 16 \cdot \cos(105^\circ)$

Calculating (\cos(105^\circ) \approx -0.2588):

$|AB|^2 = 100 + 256 + 2 \cdot 10 \cdot 16 \cdot 0.2588$

Resulting in:

$|AB|^2 = 273.1784$

Thus:

$|AB| \approx \sqrt{273.1784} \approx 16.53 m$

Therefore, | AB | = 16.53 m (to two decimal places).