Joe wants to draw a diagram of his farm - Leaving Cert Mathematics - Question 9 - 2016

To find the coordinates of point B, we add 5 units to the y-coordinate of point F. Therefore, the coordinates are:

$B = (4, 6)$

Using the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let the coordinates of Q be (Q(8, y_Q)). Then:

$d = \sqrt{(8 - 4)^2 + (y_Q - 6)^2}$

After substituting the values (assuming (y_Q = 0)):

$d = \sqrt{(4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.21$

Given that T is located directly to the east of B (point Q is aligned on the x-axis), the coordinates of T can be represented as:

$T = (8, 1)$

The area of parallelogram FBQT can be computed using the formula:

$\text{Area} = \text{base} \times \text{height}$

Where the base can be taken as the distance from F to B (5 units) and the height is the distance from F to T (which is also 4 units). Thus:

$\text{Area} = 5 \times 6 = 30 \text{ square units}$

By applying the sine rule to the triangle QFB, we have:

$\sin \angle BQF = \frac{\sin 45^\circ}{|BQ|} \cdot |FB|$

Substituting the known values,

$\angle BQF = \arcsin \left( \frac{\sin(45^\circ) \cdot |FB|}{|BQ|} \right)$

With the computations yielding:\n(\angle BQF \approx 36.9^\circ) (after rounding to one decimal place).