ABCD is a rhombus. If PB=12, AB=15, and m<ABD=24, find each measure
Heres the problem:
Thanks for your help!
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°
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°