Answer:
[tex]\mathbf{m\angle A = sin^{-1}(\frac{1}{3}) \approx 19.47^{\circ}}[/tex]
Step-by-step explanation:
This question involves some basic knowledge of trigonometric function. The following formula only works for right angled triangles.
[tex]\mathbf{sin(x) = \frac{perpendicular}{hypotenuse}}[/tex]
In ΔAKO,
m∠AKO = 90°
Therefore ΔAKO is a right angled triangle. If we take ∠A into consideration then base will be AK, perpendicular will be OK and hypotenuse will be AO.
AO = 15
OK = 5
[tex]\therefore \mathrm{sin(\angle A) = \frac{OK}{AO} = \frac{5}{15} = \frac{1}{3}}[/tex]
[tex]\mathrm{sin(\angle A)=\frac{1}{3}}[/tex]
[tex]\mathrm{\angle A=sin^{-1}(\frac{1}{3})}[/tex]
To calculate [tex]\mathrm{sin^{-1}(\frac{1}{3})}[/tex] use scientific calculator
value of [tex]\mathrm{sin^{-1}(\frac{1}{3})}[/tex] is approximately equal to 19.47°
Therefore [tex]\mathbf{m\angle A \approx 19.47}[/tex]
(NOTE : When a line passing through center of a circle, is drawn to a tangent at the point of tangency then the angle made between then is 90°
Therefore m∠AKO = 90°)
