matthew042 said:

Is there any way to do it without any trig functions like sin and cos? I don't quite understand that.

Well, you did state: "I thought relating the angles to the opposite sides would do but i was wrong."

That means using SIN ! Did you at least understand the "105" in 20 / SIN(105) ?

There is another way.

Draw a representative triangle. Then draw a perpendicular like CD with D on line AB.

You now have 2 right triangles: ACD with angles 30-60-90 and BCD with angles 45-45-90.

With a 30-60-90 triangle, the hypotenuse = twice shorter side, so AC = 2 * CD.

If you let x = BD, then CD = x, AD = 20-x and AC = 2x.

Try that.