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

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

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)

The Maximum value of a sin theta+b cos theta is:

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

The minimum value of a sin theta+b cos theta is:

If a sin theta + b cos theta = c then the value of 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

The distance between the points (a cos theta+b sin theta,0) and (0,a sin theta-b cos theta)