Question

W trojkącie abc dane są |BC|=15, sin |

Answer 1

Odpowiedź:

BC = 15 j

sin bac = 12/13

sin cba = 0,8

sin bac = h/15= 0.8

h/15= 0.8   /*15

h =  15 * 0.8

h = 12

BC^2 = h^2 + BD^2

225 = 144 + BD^2

BD^2 = 225 - 144 = 81

BD = 9 j

sin bac = 12/13

sin bac = h/CA = 12/CA

12/CA = 12/13

12/13CA = 12   /   12/13

CA = 13

CA^2 = h^2 + AD^2

13^2 = 12^2 + AD^2

169 = 144 + AD^2

AD^2 = 169 - 144 = 25

AD = 5 j

AB = 9j + 5j = 14 j

Pole =  0.5ah = 0.5 * 14 * 12 = 7 * 12 = 84 j^2