Hello!
The answer is:
![cos(R)=(1)/(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2wi3qx2t96bxi6mum3w6e2jtc0dw1vk2f0.png)
Why?
Since it's a right triangle and we need to know the adjacent side of the triangle, in order to find cos(R), we can use the Pythagorean Theorem.
Pythagorean Theorem formula:
![c^(2)=a^(2) +b^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/8v84jdw71mddd2limlr2ifnn3lbhr3syhw.png)
Where:
![c=hypotenuse=20\\a=FirstTriangleSide=AdjacentSide\\b=SecondTriangleSide=OppositeSide=10√(3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/356ahk4d1b12mayphgjndxx9qfrg4ex0g8.png)
So, the adjacent side will be:
![AdjacentSide=\sqrt{c^(2)-b^(2)}=\sqrt{20^(2)-(10√(3))^(2)}=√(400-100*3) \\AdjacentSide=√(100)=10](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wlt2a97uxfczw7ntlpf69el6qwqqtyxu1m.png)
Now, that we know the adjacent side, we can calculate cos(R), so:
![cos(R)=(AdjacentSide)/(Hypotenuse)=(10)/(20)=(1)/(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pf1g66o6z10q391wyk7nveqzteq4fy2xr8.png)
Have a nice day!