Asked by Richard
What is the value of each of the angles of a triangle whose sides are 95, 150, 190 cm in length?
I have tried to do this problem, don't get me wrong, but I don't really know where to go. Can someone tell me how to even approach this problem?
The textbook states that the answers are: 99 degrees opposite the 190 cm side; 3*10 degrees opposite the 95 cm side; 51 degrees opposite the 95 cm side.
That being said, I would still like to know how to do this problem.
I have tried to do this problem, don't get me wrong, but I don't really know where to go. Can someone tell me how to even approach this problem?
The textbook states that the answers are: 99 degrees opposite the 190 cm side; 3*10 degrees opposite the 95 cm side; 51 degrees opposite the 95 cm side.
That being said, I would still like to know how to do this problem.
Answers
Answered by
Richard
Sorry, there was a typo. *51 degrees opposite the 150 cm side.
Answered by
Damon
use law of cosines
a = 95
b = 150
c = 190
c^2 = a^2 + b^2 - 2 a b cos C
36100 = 9025 + 22500 - 28500 cos C
28500 cos C = -4575
cos C = -.16053
C = 99.2 degrees
now use law of sine to find the next one
a = 95
b = 150
c = 190
c^2 = a^2 + b^2 - 2 a b cos C
36100 = 9025 + 22500 - 28500 cos C
28500 cos C = -4575
cos C = -.16053
C = 99.2 degrees
now use law of sine to find the next one
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