1. Solve the triangle. A = 33°, a = 20, b = 13 (1 point)
[tex]\dfrac{a}{\sin A} = \dfrac{b}{\sin B}[/tex]
[tex]\sin B = \dfrac{b}{a} \sin A= \dfrac{13}{20} \sin 33 \approx 0.3540 [/tex]
[tex]B = \arcsin .3540 = 20.73^\circ[/tex]
The supplement of B (which has the same sine) is around 159° so with A exceeds 180 degrees so we ignore it.
[tex]C = 180^\circ - A - B = 180 - 33 - 20.73 = 126.27^\circ[/tex]
[tex]c = \dfrac{\sin C}{\sin A } a= \dfrac{\sin 126.27}{\sin 33} (20) \approx 29.6 [/tex]
Fourth choice: B = 20.7°, C = 126.3°, c ≈ 29.6
2. State whether the given measurements determine zero, one, or two triangles. C = 30°, a = 32, c = 16 (1 point)
[tex]\sin A = \dfrac{a}{c} \sin C= \dfrac{32}{16} \sin 30 = 2(\frac 1 2)=1[/tex]
That's unique, A=90°
One triangle.
3. Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles. A = 59°, a = 13, b = 14 (1 point)
[tex]\sin B = \dfrac{b}{a} \sin A= \dfrac{14}{13} \sin 59 \approx 0.9231[/tex]
There are two triangle angles with this sine, supplementary to each other,
[tex]B_1 \approx 67.4 \textrm{ or } B_2 \approx 180- 67.4 = 112.6[/tex]
Since when combined with A=59° neither A+B exceeds 180°, we have two valid triangles.
[tex]C_1 = 180 - 59 - 67.4 = 53.6[/tex]
[tex]c_1 = \dfrac{\sin C_1}{\sin A } a = \dfrac{\sin 53.6}{\sin 59} (13) = 12.2[/tex]
[tex]C_2 = 180 - 59 - 112.6 = 8.4[/tex]
[tex]c_2 = \dfrac{\sin C_2}{\sin A } a = \dfrac{\sin 8.4}{\sin 59} (13) = 2.2[/tex]
That's the last choice, suitably edited:
B = 67.4°, C = 53.6°, c = 12.2; B = 112.6°, C = 8.4°, c = 2.2
