Answer
No triangle satisfies the conditions given in this problem.
Work Step by Step
$a=50$ $,$ $b=100$ $,$ $\angle A=50^{\circ}$
The Law of Sines is $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$
Find the angle $B$ by using the formula $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}$, obtained from the Law of Sines. Substitute the known values into the formula and solve for $B$:
$\dfrac{\sin50^{\circ}}{50}=\dfrac{\sin B}{100}$
$\sin B=\Big(\dfrac{100}{50}\Big)\sin50^{\circ}$
$\sin B=(2)\sin50^{\circ}$
$B=\sin^{-1}[(2)\sin50^{\circ}]=Error$
$(2)\sin50^{\circ}\approx1.532$
Since $(2)\sin50^{\circ}\approx1.532$ and the sine is never greater than $1$, no triangle satisfies the conditions given in this problem.
