this question is not working out for me please help
it says solve triangle abc with the following information-
a=68
c=49
b=102 degrees
the answer is b=91.7
a=46.5 degrees
c=31.5 degrees
i keep getting over 100 for the side please help!
2007-09-25 09:21:01 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
It's a little confusing that you're using "b" for both the angle and the opposite side. Usually the convention is to use A, B, C for the angles and a, b, c for the sides.
Using the Law of Cosines, you have:
b^2 = 68^2 + 49^2 - 2(68)(49)cos(102)
I get approximately 91.7 units for b too.
Are you saying you get something bigger?
2007-09-25 09:26:54 · answer #1 · answered by Anonymous · 0⤊ 0⤋
a=68
c=49
B=102 degrees Use capital letters for angles.
b^2 = a^2+c^2 -2accos(102) = 68^2 +49^2 - 2(68)(49)(cos(102)) = 4624 + 2401 +1385.52 = 8410.52
b= sqrt(8410.52) = 91.71
c/sinC = 91.71/sin(102) --> C = arcsin49/93.76= 31.51 degrees. So A = 180-102-31.51 = =46.49 degrees
2007-09-25 09:41:18 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
b^2 = a^2 + c^2 - 2ac cos(B)
= 68^2 + 49^2 - 2(68)(49)cos(102)
= 4624 + 2401 - 6664 (-.2079)
= 7025 + 1385.523
= 8410.523
b = 91.7
2007-09-25 09:31:53 · answer #3 · answered by PeterT 5 · 0⤊ 0⤋