Vijayan Sir , both AP and BQ are perpendicular to XY , so AP is parallel to BQ. If we construct a ray from Q parallel to AB which intersect AP at say S , then triangle PQS will be a right angle triangle with angle P = 90°(Radius is perpendicular to tangent) , from there we can apply Pythagoras Theorem in triangle PQS.

pl correct the qn no.2(ptolemy,s theorem)

ReplyDeleteAB.CD+AD.BC=AC.BD (not AC.AD)

pl explain the second step of third qn.(PQ^2=AB^2-(a-b)^2)

ReplyDeleteVijayan Sir , both AP and BQ are perpendicular to XY , so AP is parallel to BQ. If we construct a ray from Q parallel to AB which intersect AP at say S , then triangle PQS will be a right angle triangle with angle P = 90°(Radius is perpendicular to tangent) , from there we can apply Pythagoras Theorem in triangle PQS.

ReplyDeleteVijayan Sir , corrected the mistake. Thank you.

ReplyDelete