## Trigonometry (11th Edition) Clone

There are no values of $b$ where angle $A$ has two possible solutions.
Since $B = 150^{\circ}$, then $0 \lt A \lt 30^{\circ}$, otherwise, no such triangle exists. We can use the law of sines to determine the values for $b$: $\frac{b}{sin~B} = \frac{a}{sin~A}$ $sin~A = \frac{a~sin~B}{b}$ $sin~A = \frac{10~sin~150^{\circ}}{b}$ $sin~A = \frac{5}{b}$ Since $sin~30^{\circ} = 0.5$, then $sin~A \lt 0.5$ $\frac{5}{b} \lt 0.5$ $b \gt 10$ If $b \gt 10$, then $0 \lt A \lt 30^{\circ}$ and the angle $A$ has exactly one value. If $0 \lt b \leq 10$, then $sin~A \gt 0.5$ and the angle $A$ has no solutions such that $0 \lt A \lt 30^{\circ}$ Therefore, there are no values of $b$ where angle $A$ has two possible solutions.