Joe wants to draw a diagram of his farm - Leaving Cert Mathematics - Question 9 - 2018

Since point B is 5 units directly north of point F, the coordinates of point B will be:

$B = (4, 6)$

This point should be plotted and labeled accordingly on the diagram.

To find the distance from point B to point Q, we can use the distance formula:

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

Where:

• $(x_1, y_1)$ are the coordinates of B, which are $(4, 6)$
• $(x_2, y_2)$ are the coordinates of Q, which are $(-2, 7)$

Substituting these values into the formula gives:

$d = ext{sqrt}((-2 - 4)^2 + (7 - 6)^2)$ $d = ext{sqrt}((-6)^2 + (1)^2)$

$$d ext{ is approximately } 6.08 ext{ units (to 2 decimal places).}

Given that FBQT is a parallelogram, the coordinates of point T can be derived from the relationship of the coordinates of F and B. Thus:

$T = (-2, 2)$

The area A of a parallelogram can be calculated using:

$A = base \times height$

Here, the base is the length from B to F (which is 5 units) and the height can be derived from the vertical distance from T to F (which is also 6 units). Hence:

$A = 5 \times 6 = 30 ext{ square units.}$

Using the sine rule:

$\frac{sin(\angle LQBF)}{|FB|} = \frac{sin(45°)}{|QB|}$

Given that |FB| = 5 units and |QB| = 6.08 units, the calculations yield:

$\sin(\angle LQBF) = \frac{5 \cdot sin(45°)}{6.08}$

Calculating gives:

$\angle LQBF \approx 35.1° ext{ (to one decimal place).}$