Given the figure, we can deduce the following information:
Angle C= 80°
AC=565 ft
BC=480 ft
To determine the area of the triangle ABC and the distance AB, we first
redraw the figure as shown below:
Next, we use the formula:
[tex]c=\sqrt{a^2+b^2-2abcosC}[/tex]
where:
a=BC=480 ft
b=AC=565 ft
C= Angle C=80°
c=AB
We plug in what we know:
[tex]\begin{gathered} c=\sqrt{a^2+b^2-2abcosC} \\ c=\sqrt{480^2+565^2-2(480)(565)cos80} \\ Calculate \\ c=AB=674.86\text{ }ft \end{gathered}[/tex]
Then, we get the area by using the formula:
[tex]A=\frac{absin\theta}{2}[/tex]
where:
a=480
b=565
θ=80°
So,
[tex]\begin{gathered} A=\frac{abs\imaginaryI n\theta}{2} \\ A=\frac{(480)(565)sin80}{2} \\ Calculate \\ A=133539.93\text{ }ft^2 \end{gathered}[/tex]
Therefore, the answers are:
Distance AB=674.86 ft
Area of triangle ABC=133539.93 ft^2
