## Thomas' Calculus 13th Edition

$\approx 42.22^o$
We know that the three angles spanning the line must add up to 180 degrees: $65^o (90^o-\beta)+(90^o-\alpha)=180^o$ Solving for $\alpha$, we get: $\alpha=65^o-\beta=65^o-tan^{-1}(\frac{21}{50})$ Which approximately equals: $\approx65^o - 22.78^o\approx42.22^o$