Answer:
[tex]A\approx 51.96[/tex]
Step-by-step explanation:
First, we can solve for the measure of angle C using the Law of Sines:
[tex]\dfrac{\sin(A)}{a}=\dfrac{\sin(B)}{b} = \dfrac{\sin(C)}{c}[/tex]
where [tex]A[/tex] is the angle opposite side [tex]a[/tex].
↓ plugging in the given values
[tex]\dfrac{\sin(C)}{13} = \dfrac{\sin(31\°)}{8}[/tex]
↓ multiplying both sides by 13
[tex]\sin(C) = \dfrac{13\sin(31\°)}{8}[/tex]
↓ taking the inverse sine of both sides
[tex]C = \sin^{-1}\!\left(\dfrac{13\sin(31\°)}{8}\right)[/tex]
↓ evaluating using a calculator
[tex]C\approx 56.82\°[/tex]
We can now solve for the measure of angle A using our knowledge that the interior angles of a triangle sum to 180°:
[tex]A+B+C=180\°[/tex]
↓ plugging in the known values
[tex]A + 31\° + 56.82\° = 180\°[/tex]
↓ subtracting [tex](31\° + 56.82\°)[/tex] from both sides
[tex]A = 92.18\°[/tex]
Next, we can again use the Law of Sines to solve for the length of side a:
[tex]\dfrac{\sin(92.18\°)}{a}=\dfrac{\sin(31\°)}{8}[/tex]
↓ taking the reciprocal of both sides
[tex]\dfrac{a}{\sin(92.18\°)}=\dfrac{8}{\sin(31\°)}[/tex]
↓ multiplying both sides by [tex]\sin(92.18\°)[/tex]
[tex]{a}=\dfrac{8\sin(92.18\°)}{\sin(31\°)}[/tex]
↓ evaluating using a calculator
[tex]a\approx 15.52[/tex]
Finally, now that we know all of the triangle's side lengths, we can solve for its area using Heron's formula:
[tex]A=\sqrt{s(s-a)(s-b)(s-c)}[/tex]
where [tex]s[/tex] is the triangle's semi-perimeter, or half-perimeter.
We know:
- [tex]a=15.52[/tex]
- [tex]b=8[/tex]
- [tex]c=13[/tex]
Hence,
- [tex]s=\dfrac{15.52 + 8 + 13}{2} = 18.26[/tex]
↓ plugging the known values into the area formula
[tex]A=\sqrt{18.26(18.26-15.52)(18.26-8)(18.26-13)}[/tex]
[tex]\boxed{A\approx 51.96}[/tex]
