The given triangle is a right-angled triangle.
Consider PO is Hypotenuse.
By using the Pythagoras formula, we get
[tex]PO^2=8^2+6^2[/tex]
[tex]PO^2=64+36[/tex][tex]PO^2=100=10^2[/tex][tex]PO=10[/tex]
Consider the angle x:
Recall the sine formula
[tex]\sin \theta=\frac{Opposite\text{ side}}{\text{Hypotenuse}}[/tex]
Substitute Opposite side =6 and Hypotenuse=10, we get
[tex]\sin x^o=\frac{6}{10}[/tex]
[tex]\sin x^o=\frac{3}{5}[/tex][tex]\text{Use sin }36.869=\frac{3}{5}[/tex]
[tex]\sin x^o=\sin 36.869[/tex][tex]x^o=37[/tex]
Consider the angle y:
Recall the sine formula
[tex]\sin \theta=\frac{Opposite\text{ side}}{\text{Hypotenuse}}[/tex]
Substitute Opposite side =8 and Hypotenuse=10, we get
[tex]\sin y^o=\frac{8}{10}[/tex]
[tex]\sin y^o=\frac{4}{5}[/tex]
[tex]\text{Use }\sin 53.13^{}=\frac{4}{5}[/tex]
[tex]\sin y^o=\sin \text{ 53.13}[/tex][tex]y^o=53^{}[/tex]
Hence the required values are
[tex]x^o=37^o[/tex][tex]y^o=53^o[/tex]
