Question

Plz hurry!!!! thank you!!!!

Answer 1

Answer:

∠A ≈ 19.47°

Step-by-step explanation:

As you know , the angle that tangent makes with the radius at the point of tangency is 90°.

∴∠OKA = 90°

As  AO = 15 and OK = 5 ,

$sin(A) = \frac{OK}{AO}$

$sin(A) = \frac{5}{15}$

$sin(A) = \frac{1}{3}$

∠A = $sin^{-1} (\frac{1}{3} )$

∠A ≈ 19.47°

Answer 2

Answer:

$\mathbf{m\angle A = sin^{-1}(\frac{1}{3}) \approx 19.47^{\circ}}$

Step-by-step explanation:

This question involves some basic knowledge of trigonometric function. The following formula only works for right angled triangles.

$\mathbf{sin(x) = \frac{perpendicular}{hypotenuse}}$

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

$\therefore \mathrm{sin(\angle A) = \frac{OK}{AO} = \frac{5}{15} = \frac{1}{3}}$

$\mathrm{sin(\angle A)=\frac{1}{3}}$

$\mathrm{\angle A=sin^{-1}(\frac{1}{3})}$

To calculate $\mathrm{sin^{-1}(\frac{1}{3})}$ use scientific calculator

value of $\mathrm{sin^{-1}(\frac{1}{3})}$ is approximately equal to 19.47°

Therefore $\mathbf{m\angle A \approx 19.47}$

(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°)