In 22, you're looking for the vertical height of the triangle. You're given the angle opposite the side you want to find (which I'll call [tex]x[/tex]) and the length of the hypotenuse. This sets you up with the relation
[tex]\sin2^\circ=\dfrac x{1600}\implies x\approx55.84\text{ m}[/tex]
In 23, you're given a similar situation, except now you're looking for the angle (I'll call it [tex]\theta[/tex]) in the triangle opposite the side denoting the height of the airplane. So this time,
[tex]\sin\theta=\dfrac{1500}{5000}=\dfrac3{10}\implies\theta=\arcsin\dfrac3{10}\approx17.46^\circ[/tex]
