Answer:
Givens
[tex]m \angle BAC=63\°[/tex]
[tex]m(BE)=96\°[/tex]
[tex]DE[/tex] is secant.
[tex]DG[/tex] is tangent.
(A)
Remember that the arc subtended by a central angle is equal to it. That means
[tex]m(BC)=m\angle BAC = 63\°[/tex]
(B)
Now, by sum of arcs, we know
[tex]m(EBC)=m(EB)+m(BC)=96\° +63\°=159\°[/tex]
(C)
By definition, the total arc of a circle is 360°, so
[tex]m(EFC)=360\°-m(EBC)=360\° - 159\°=201\°[/tex]
(D)
We know that the external angle formed by a secant and a tanget is one half of the difference between the intercepted arcs, so
[tex]\angle D=\frac{1}{2}(m(EFC)-m(BC))=\frac{1}{2} (201\° - 63\°)=\frac{1}{2} (138\°)=69\°[/tex]
So, the angle D is 69°.
