1) In triangle ABC, we know that AB = 5, BC = 9, and CA = 7. D is a point on BC such that AD bisects ∠BAC. What is DC? 2)] O is the circumcenter of 4ABC. If OB = OC = BC, what is ∠BAC

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  1) 5.25

  2) 95.74°

Step-by-step explanation:

1) The angle bisector divides the triangle into proportional segments.

  DC/BD = CA/BA

We know that BD+DC = 9, so we can fill in the above equation with known values to get ...

  DC/(9-DC) = 7/5

  5·DC = 7(9 -DC) . . . . cross multiply

  12·DC = 63 . . . . . add 7·DC

  DC = 63/12 = 5.25

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2) Angle BAC can be found using the Law of Cosines.

  a² = b² +c² -2bc·cos(A)

  cos(A) = (b² +c² -a²)/(2bc) = (7² +5² -9²)/(2·7·5) = -7/(70) = -1/10

  A = arccos(-1/10) ≈ 95.739°

The measure of ∠BAC is about 95.74°.

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Additional comment

The circumcircle is irrelevant to the questions asked here.

7.3

step-by-step explanation: