ABCD is a rhombus. If PB=12, AB=15, and m<ABD=24, find each measure

Heres the problem:

Thanks for your help!

Guest Apr 5, 2015

#1**+10 **

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°

CPhill Apr 5, 2015

#1**+10 **

Best Answer

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°

CPhill Apr 5, 2015