If the segment joining the points A(a,b)  and  B(c,d)  subtends an angle θ  at the origin, then

Let O be the origin. Then OA2=a2+b2,OB2=c2+d2  and AB2=(c−a)2+(d−b)2 Using cosine formula in ΔOAB,  we haveAB2=OA2+OB2−2.OA.OBcosθ⇒   (c−a2)+(d−b)2=a2+b2+c2+d2−2a2+b2c2+d2cosθ⇒ c2+a2−2ac+d2+b2−2bd=a2+b2+c2+d2−2a2+b2c2+d2cosθ⇒ 2(ac+bd)=2a2+b2c2+d2cosθ⇒   cosθ=ac+bda2+b2c2+d2.