Given:
m∠B = 45°
b = 4
c = 5
From the Law of Sines, obtain
[tex] \frac{sinC}{c}= \frac{sinB}{b} \\ sinC=( \frac{c}{b})sinB \\sinC = ( \frac{5}{4} )sin(45^{o})=0.884\\ C = sin^{-1}0.884=62.1^{o}[/tex]
This yields
m∠A = 180 - 45 - 62.1 = 72.9°
[tex]a=( \frac{sinA}{sinB})b=( \frac{sin(72.9^{o})}{sin(42^{o})})4=5.41[/tex]
The first triangle has
∠A=72.9°, m∠B=45°, m∠C = 62.1°, a=5.41, b=4, c=5.
Also,
[tex]m\angle{C} = sin^{-1}0.884 = 117.9^{o}[/tex]
This yields
m∠A = 180 - 45 - 117.9 = 17.1°
[tex]a=( \frac{sinA}{sin(45^{o})} )4=1.66[/tex]
The second triangle has
m∠A = 17.1°, m∠B = 45°, m∠C = 117.9°, a = 1.66, b = 4, c = 5
Answer: There are 2 possible triangles.
