Answer:
![Sin(\alpha)=(2)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vego7r1q4ybv0ib2chk6sd7rwafbmzamve.png)
Explanation:
We need a simple identity to solve for sine of an obtuse angle (angle greater than 90 degrees).
We know,
![Sin(180-\alpha)=\alpha](https://img.qammunity.org/2020/formulas/mathematics/middle-school/fznljwknqojcdftb8zhpvjza9ymwyzzpa0.png)
So the value of
would depend on the triangle we can make on the left side of the coordinate system shown.
The triangle would be as shown in the attached figure.
This triangle has base length of
and height of 2. The hypotenuse, r, can be solve using pythagorean theorem:
![(√(5) )^2+(2)^2=r^2\\5+4=r^2\\9=r^2\\r=3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/i7wprmc6y18odjgzx5cyytkgrfb0u3g2qw.png)
We know sin of an angle is "opposite" side over "hypotenuse". The triangle's opposite is "2" and hypotenuse is "3". So we can finally write:
![Sin(\alpha)=(Opposite)/(Hypotenuse)=(2)/(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/tojiyqk4avfg91afsc4msrfunpysbgc6x6.png)