Question

In circle A shown, secant DE and tangent DG are drawn. It ir know that m BAC = 63 and mBE = 96.

Answer 1

Answer:

Givens

$m \angle BAC=63\°$

$m(BE)=96\°$

$DE$ is secant.

$DG$ is tangent.

(A)

Remember that the arc subtended by a central angle is equal to it. That means

$m(BC)=m\angle BAC = 63\°$

(B)

Now, by sum of arcs, we know

$m(EBC)=m(EB)+m(BC)=96\° +63\°=159\°$

(C)

By definition, the total arc of a circle is 360°, so

$m(EFC)=360\°-m(EBC)=360\° - 159\°=201\°$

(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

$\angle D=\frac{1}{2}(m(EFC)-m(BC))=\frac{1}{2} (201\° - 63\°)=\frac{1}{2} (138\°)=69\°$

So, the angle D is 69°.