Let O be the circumcentre of an obtuse triangle ABC with A = 15^ and B = 45^ . If the position vector of C with respect to O can be expressed as OC = OA + OB , then the value of 3 ( + ) is equal to

Step-by-step solution

In ABC , the sum of angles is 180^ . C = 180^ - (15^ + 45^ ) = 120^ . Since O is the circumcentre, the angle subtended by a chord at the centre is twice the angle subtended at the circumference. For the obtuse angle C , the central angle for the minor arc AB is 2(180^ - 120^ ) = 120^ . Thus, AOB = 120^ . Similarly, BOC = 2 A = 30^ and AOC = 2 B = 90^ . Let the circumradius be R . Then | OA | = | OB | = | OC | = R . The dot products are: OA OB = R^2 120^ = - R^2 2 OB OC = R^2 30^ = 3 R^2 2 OA OC = R^2 90^ = 0 Given