2017-II-21
Ans: C
Join $AC$.
… Read Join $AC$.
![](https://solvingmaster.space/wp-content/uploads/2021/06/2017PaperII21_fig01-300x278.png)
In quadrilateral $ABCD$,
$\begin{array}
In quadrilateral $ABCD$,
$\begin{array}
$\begin{array}{rcl}
\cos 40^\circ
$\begin{arr
$\begin{array}{cl}
=
Since $DE$ is the tangent to the circle at $A