1. ## 3 circle questions

in attachments

2. Jelo, dgenerationx2!

For the first two, we need this theorem:

The angle between a tangent and a secant (to the point of tangency)
. . is one-half the intercepted arc.

1) Given: . $\text{arc}(AC) \,=\,50^o,\;\;\text{arc}(AFD) \,=\,190^o,\;\;\text{tangent } BE.$ . Find: $\angle BAC.$

. . $(A)\;25^o\qquad (B)\;50^o\qquad(C)\;95^o \qquad (D)\;60^o$
Code:
         D    * * *
o           *
*  *            o
*               F *
*
*                   * 190°
*      *            *
C o                   *
*     *
*  *              *
50° *   *  *        *
*   *       *
B - - - * o * - - - - E
A

$\text{arc}(AFD) = 190^o,\;\text{arc}(AC) = 50^o \quad\Rightarrow\quad \text{arc}(CD) = 120^o \quad\Rightarrow\quad \angle CAD \,=\,60^o$

$\text{arc}(DCA) = 170^o \quad\Rightarrow\quad \angle DAB \,=\,85^o$

Therefore: . $\angle BAC \:=\:85^o-60^o \:=\:25^o$ . . . answer (A)

Given: .tangent $AB$ and secant $BC,\;\;\text{arc}(BDC) = 200^o$ . Find: $\angle ABC$
Code:
              * * *
*           *
*               *  C
o D                o
/
*                /  *
200° *               /   *
*              /    *
/
*           /     *
*         /     *
*      /    *
* o * - - - - - A
B

We hafve: . $\text{arc}(BDC) = 200^o \quad\Rightarrow\quad \text{arc}(BC) = 160^o$

Therefore: . $\angle ABC \,=\,80^o$ . . . answer (A)