From a fixed point P on the circumference of a circle of radius a the perpendicular PR is drawn to the tangent at Q [a variable point] , then the maximum area of  TRIANGLE PQR ....IS??

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this is the answer of this problem

To make the calculation simple,we assume centre of circle to be origin.Let P=(acosp,bsinp) be the fixed point on the circumference and Q be the variable point (acosk,bsink)

then,eqn of tangent to the circle at Q is xcosk+ysink=a

PQ,which is the hypotenuse of the triangle comes out to be 2asin(p-k)/2.PR,the perpendicular distance of R from the tangent comes out as 2asin²(p-k)/2.this gives QR=2asin(p-k)/2cos(p-k)/2

area=2a²sin³(p-k)/2cos(p-k)/2

TO obtain max area,differentiate w.r.t. k,which gives tan(p-k)/2=rt3  or  (p-k)/2=60⁰

so,the max area is 3rt3a²/8

hey kanika you chose wrong field this is differential calculus not analytical maths.