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. d. 64

8. i. 32

9. c. 125

10. j. none of the above

11. d. 81

12. i. 216

13. e. none of the above

14. g. 240

15. e. none of the above

16. g. 32

17.   e. none of the above

597

step-by-step explanation: