Trigonometriske beregninger af trekanter

Opgave 1

= 60°

a = 8.24 c = 16

= 90° b = 13,71 °B = 180-30-90=60°a = 16*sin(31°)=8,24

b = 90*cos(31°)=13,71

Opgave 2

ABC

A

B

a

b

c

1

29°

61°

1.35

2.44

2,8

2

45.8

44,2°

28,67

27,88

40

3

50,17°

39,82°

22,11

18,44

28,79

4

34,9°

55,09°

5,78

8,28

10,10

5

43,4°

46,6°

12

12,68

17,46

6

17,8°

72,2°

1,63

5,1

5,35

7

6°

84,0°

5,92

56,32

56,63

a = 2.8*sin(29°)=1.35b = 2.8*cos(29°)=2.44Vinklen B = 180-90-29=61°

a = 40*cos(44,2)=28,67b = 40*sin(44.2)=27,88A = 180-90-44,2=45.8°

c = Kvadratroden af(22,112+18,442)=28.79A = tan-1(22.11/18.44)=50,17°B = tan-1(18,44/22,11)=39,82°

b = ?(10,102-5,782)=8,28A = sin-1(5,78/10,10)=34,9°B = cos-1(5,78/10,10)=55,09°

b = 12/tan(43,4)=12,68c = 12/sin(43,4)=17,46B = 180-90-43,4=46,6°

a = 5,1*tan(17,8)=1,63c = 5,1/cos(17,8)=5,35B = 180-90-17,8=72,2°

b = 5,92*tan(84)=56,32

c = 5,92/cos(84)=56,63A = 180-90-84=6°

Opgave 3

AC = 10 fordi den er lige så lang som CD.

B =

BD = 40 - regnet via. aflæsning, da vi ved CD er 10, er det nemt at se BD er 40.