Answer:
[tex]m\angle B\approx 80.27^\circ\text{ and } m\angle C\approx 59.73^\circ[/tex]
Side AB measures approximately 20.15 units.
Step-by-step explanation:
We can use the Law of Sines, given by:
[tex]\displaystyle \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}[/tex]
Since we already know ∠A and a, we can use it to determine our other measures.
We know that b is 23. Substitute:
[tex]\displaystyle \frac{\sin 40^\circ}{15}=\frac{\sin B}{23}[/tex]
Solve for ∠B:
[tex]\displaystyle \sin B=\frac{23\sin 40^\circ}{15}[/tex]
We can take the inverse sine of both sides:
[tex]\displaystyle m\angle B=\sin^{-1}\frac{23\sin 40^\circ}{15}[/tex]
Use a calculator (make sure you're in Degrees mode!):
[tex]m\angle B\approx 80.27^\circ[/tex]
The interior angles of a triangle must total 180°. Hence:
[tex]m\angle A+m\angle B+m\angle C=180^\circ[/tex]
Substitute in known values:
[tex](40^\circ)+(80.27^\circ)+m\angle C=180^\circ[/tex]
Therefore:
[tex]m\angle C\approx 59.73^\circ[/tex]
Lastly, to find AB or c, we can use the Law of Cosines:
[tex]c^2=a^2+b^2-2ab\cos(C)[/tex]
a = 15, b = 23, and to prevent rounding errors, we will use the exact value for C. Recall that we know that:
[tex]m\angle A+m\angle B+m\angle C=180^\circ[/tex]
Using the exact value for ∠B, we acquire:
[tex]40^\circ+\displaystyle \sin^{-1}\frac{23\sin40^\circ}{15}+m\angle C=180^\circ[/tex]
Therefore:
[tex]\displaystyle m\angle C=\left(140-\sin^{-1}\frac{23\sin40}{15}\right)^\circ[/tex]
Substitute:
[tex]\displaystyle c^2=(15)^2+(23)^2-2(15)(23)\cos\left(140-\sin^{-1}\frac{23\sin40^\circ}{15}\right)[/tex]
Simplify and take the square root of both sides.
[tex]\displaystyle c=\sqrt{754-690\cos\left(140-\sin^{-1}\frac{23\sin40^\circ}{15}\right)}[/tex]
And use a calculator:
[tex]c\approx 20.15[/tex]
