Answer:
![Sin \theta = (Perpendicular)/(Hypotenuse)=(3)/(√(13))](https://img.qammunity.org/2021/formulas/mathematics/high-school/t38po4767qpf67z710ya2bm1z8ow3actmo.png)
![Cos \theta = (Base)/(Hypotenuse)=(2)/(√(13))](https://img.qammunity.org/2021/formulas/mathematics/high-school/qtwxick2jddqqpfzhthd4e7ugxrwgr8bco.png)
![Tan \theta = (Perpendicular)/(Base)=(3)/(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/cspmsapehw0c0eias3i9o0qzspd3xrbtb3.png)
Explanation:
We are given that The point (2, 3) is on the terminal side of angle Θ, in standard position
First Draw a vertical line from the point(2,3) to the x axis.
So, Length of vertical line is 3
The intersection of the line with the x axis is at x=2.
So, now we have obtained a triangle with the horizontal side of length 2, the vertical side of length 3
To Find hypotenuse we will use Pythagoras theorem
![Hypotenuse^2=Perpendicular^2+Base^2\\Hypotenuse^2=3^2+2^2\\Hypotenuse=√(9+4)\\Hypotenuse=√(13)](https://img.qammunity.org/2021/formulas/mathematics/high-school/r3q1xzadebu11e3tf73enhg097kssp9raf.png)
![Sin \theta = (Perpendicular)/(Hypotenuse)=(3)/(√(13))](https://img.qammunity.org/2021/formulas/mathematics/high-school/t38po4767qpf67z710ya2bm1z8ow3actmo.png)
![Cos \theta = (Base)/(Hypotenuse)=(2)/(√(13))](https://img.qammunity.org/2021/formulas/mathematics/high-school/qtwxick2jddqqpfzhthd4e7ugxrwgr8bco.png)
![Tan \theta = (Perpendicular)/(Base)=(3)/(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/cspmsapehw0c0eias3i9o0qzspd3xrbtb3.png)