Parallelograms

23.  Using the Law of Cosines, we have

AP^2  = 15^2 + 12^2  - 2(15)(12)cos 24

AP = about 6.33

24. The diagonals of a rhombus bisect each other, therefore AP = PC  = 6.33

25. Since the diagonals bisect each other, BP = PD = 12. So, BD =24. And we can use the Law of Cosines to find AD

AD^2  = 24^2 + 15^2 - 2(24)(15)cos 24

AD  = about 11.97

So, using the Law of Sines

sin BDA / AB = sin 24 / AD

sinBDA / 15 = sin 24 / 11.97

sin^-1(15 sin 24 / 11.97) = BDA = 30.64°

26. We can use the Law of Cosines to find ACB....note BC = AD = 11.97  and AC = 2(AP) =2(6.33) = 12.66

AB^2 = BC^2 + AC^2 - 2 - 2(BC)(AC)cosACB

15^2 = 11.97^2 + 12.66^2 - 2(11.97)(12.66)cosACB

cos^-1 = (15^2 - 11.97^2 - 12.66^2) / (-2(11.97)(12.66)) = ACB = 74.98°

Not sure how to give CPhill's answer a vote or thumbs up, but thank you, very helpful! 

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