Answer
No triangle satisfies the conditions given.
Work Step by Step
$a=50$ $,$ $b=65$ $,$ $\angle A=55^{\circ}$
Find angle $B$ by using the formula $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}$, obtained from the Law of Sines. Substitute the known values and solve for $B$:
$\dfrac{\sin55^{\circ}}{50}=\dfrac{\sin B}{65}$
$\sin B=\Big(\dfrac{65}{50}\Big)\sin55^{\circ}$
$\sin B=\Big(\dfrac{13}{10}\Big)\sin55^{\circ}$
$B=\sin^{-1}\Big[\Big(\dfrac{13}{10}\Big)\sin55^{\circ}\Big]$
Since $\Big(\dfrac{13}{10}\Big)\sin55^{\circ}\approx1.0649$ and the sine is never greater than one, no triangle satisfies the conditions given.
