Question

the price of a fenced property
a triangular piece of land is represented by the triangle abc on the right.
we have:
$moverline{ab} = 50$ m, $mangle b = 62^circ$ and $mangle c = 38^circ$.
the land on this property sells for $40 per square metre, and the fence, around the property, sells for $8 per metre.
what is the total selling price, to the nearest dollar, of this fenced property?

Explanation:

Step1: Calculate ∠A

The sum of angles in a triangle is $180^\circ$.
$\angle A = 180^\circ - 62^\circ - 38^\circ = 80^\circ$

Step2: Find length of BC

Use the Law of Sines: $\frac{AB}{\sin C} = \frac{BC}{\sin A}$
$BC = \frac{AB \cdot \sin A}{\sin C} = \frac{50 \cdot \sin 80^\circ}{\sin 38^\circ} \approx \frac{50 \cdot 0.9848}{0.6157} \approx 80.03$ m

Step3: Find length of AC

Use the Law of Sines: $\frac{AB}{\sin C} = \frac{AC}{\sin B}$
$AC = \frac{AB \cdot \sin B}{\sin C} = \frac{50 \cdot \sin 62^\circ}{\sin 38^\circ} \approx \frac{50 \cdot 0.8829}{0.6157} \approx 71.61$ m

Step4: Calculate fence cost

First find perimeter: $P = AB + BC + AC \approx 50 + 80.03 + 71.61 = 201.64$ m
Fence cost = $8 \times 201.64 \approx 1613.12$ dollars

Step5: Calculate triangle area

Use formula: $Area = \frac{1}{2} \cdot AB \cdot BC \cdot \sin B$
$Area \approx \frac{1}{2} \times 50 \times 80.03 \times \sin 62^\circ \approx 250 \times 80.03 \times 0.8829 \approx 1766.47$ m²

Step6: Calculate land cost

Land cost = $40 \times 1766.47 \approx 70658.8$ dollars

Step7: Calculate total cost

Total cost = Fence cost + Land cost
$1613.12 + 70658.8 \approx 72271.92$

Answer:

$\$72272$