Draw right-angled triangle ABC, where B is 90 degrees. Let A be theta then cos(theta)=7/25 where hypotenuse AC=25 and AB=7. By Pythagoras, BC^2=AC^2-AB^2=625-49=576, BC=sqrt(576)=24. Therefore sin(theta)=BC/AC=24/25. This is the QI (quadrant 1) solution. If the triangle is constructed in QIV (quadrant 4), B is (0,0), A is (7,0) and C is (0,-24). cosBAC=cos(theta)=AB/AC=7/25; sin(theta)=BC/AC=-24/25, because BC is in the negative part of the quadrant.