Right away, we know that
[tex]\cos\beta=\dfrac{16}{20}=\dfrac45[/tex]
which means
[tex]\sin\beta=\sqrt{1-\cos^2\beta}=\dfrac35[/tex]
Then
[tex]\cos\beta=\cos2\left(\dfrac\beta2\right)=\cos^2\dfrac\beta2-\sin^2\dfrac\beta2[/tex]
[tex]\sin\beta=\sin2\left(\dfrac\beta2\right)=2\sin\dfrac\beta2\cos\dfrac\beta2[/tex]
Let [tex]x=\cos\dfrac\beta2[/tex] and [tex]y=\sin\dfrac\beta2[/tex]. Then
[tex]\begin{cases}\dfrac45=x^2-y^2\\\\\dfrac35=2xy\end{cases}\implies\begin{cases}\cos\dfrac\beta2=\dfrac3{\sqrt{10}}\\\\\sin\dfrac\beta2=\dfrac1{\sqrt{10}}\end{cases}[/tex]
from which we find
[tex]\tan\dfrac\beta2=\dfrac{\sin\frac\beta2}{\cos\frac\beta2}=\dfrac13[/tex]
You can use the same ideas above to find the trig ratios for [tex]\dfrac\alpha2[/tex], starting from the given fact that [tex]\sin\alpha=\dfrac{16}{20}=\dfrac45[/tex].
