Answer
No triangle satisfies the conditions given in this problem.
Work Step by Step
$a=20$ $,$ $c=45$ $,$ $\angle A=125^{\circ}$
The Law of Sines is $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$
Find the angle $C$ using the formula $\dfrac{\sin A}{a}=\dfrac{\sin C}{c}$, obtained from the Law of Sines. Substitute the known values and solve for $\angle C$:
$\dfrac{\sin 125^{\circ}}{20}=\dfrac{\sin C}{45}$
$\sin C=\Big(\dfrac{45}{20}\Big)\sin125^{\circ}$
$C=\sin^{-1}\Big[\Big(\dfrac{45}{20}\Big)\sin125^{\circ}\Big]=Error$
$\Big(\dfrac{45}{20}\Big)\sin125^{\circ}\approx1.843$
Since $\Big(\dfrac{45}{20}\Big)\sin125^{\circ}\approx1.843$ and the sine is never greater that $1$, no triangle satisfies the conditions given in this problem.
