We will solve it using the trigonometric ratio i.e. SOH
[tex] \boxed{ \sf \: \sin( \theta) = \frac{opposite}{hypotenuse} }[/tex]
In our case,
- Opposite = 20
- Hypotenuse = 25
[tex] \tt \sin( \theta) = \frac{20}{25} [/tex]
[tex] \tt \sin( \theta) = \frac{4}{5} [/tex]
[tex] \tt \theta = \arcsin( \frac{4}{5} )[/tex]
[tex] \tt \theta = 53.1 \degree[/tex]
- ∠A = 53° (when rounded off)
