The area of the triangle with vertices `(a, 0), (a cos theta, b sin theta), (a cos theta, -b sin theta)` is

A : The maximum area of the triangle formed by the points (0, 0), (a cos theta, b sin theta), (a cos theta, -b sin theta) is (1)/(2)|ab| . R : Maximum value of sin theta is 1.

The maximum area of the triangle formed by the points (0,0)*(a cos theta,b sin theta) and (a cos theta,-b sin theta) (in square units)

If a gt 0, b gt 0 then the maximum area of the parallelogram whose three vertices are O(0,0), A(a cos theta, b sin theta) and B(a cos theta, -b sin theta) is

Area of a triangle whose vertices are (a cos theta, b sin theta) , ( - a sin theta , b cos theta) " and " ( - a cos theta, - b sin theta) is

a cos theta+b sin theta+c=0a cos theta+b_(1)sin theta+c_(1)=0

The centroid of a triangle formed by the points (0, 0), (cos theta, sin theta) and (sin theta, -cos theta) lies on the line y=2x . Then theta is :

If a cos theta - b sin theta = c then a sin theta + b cos theta=