3 circle questions

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.

Quote:

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)

Quote:

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)