2017-II-13
Ans: B
$\begin{array}{rcl}
T(1) & = & 1 \\
T(2)
… Read $\begin{array}{rcl}
T(1) & = & 1 \\
T(2)
$\
Since $\Delta CEH \sim \Delta CBG$, then we have
$\beg
Method 1:
In $\Delta ACD$ and $\Delta BDE$,
$\angle CA
In quadrilateral $ABCD$,
$\begin{array}
$\begin{array}{rcl}
\cos 40^\circ
$\begin{arr
$\begin{array}{cl}
=