2015-II-18

16-06-2021 app.cch No Comments

Ans: A
Consider

Δ A B C

, we have

$\begin{array}{rcl} \tan β & = & \frac{A C}{A B} \\ A C & = & A B \tan β \end{array}$

Consider $Δ A C D$ , we have

$\begin{array}{rcl} \cos α & = & \frac{A D}{A C} \\ \cos α & = & \frac{A D}{A B \tan β} \\ \frac{A D}{A B} & = & \cos α \tan β \end{array}$