How would I solve for ALL missing sides and angles in these triangles using cosines law?:
XYZ: x=29m, y=15m,
GHI: g=13cm, h=8cm, i=15cm
Thanks a lot. Sorry for the many math questions, I find math somewhat difficult :/
XYZ: x=29m, y=15m,
GHI: g=13cm, h=8cm, i=15cm
Thanks a lot. Sorry for the many math questions, I find math somewhat difficult :/
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z² = x² + y² - 2xycos(Z)
z² = 29² + 15² - 2(29)(15)cos(122) = 841 + 225 - (870)cos(122) ≈ 1527
z ≈ 39
I would recommend using the law of sines to find one of the remaining angles but, if you insist on the law of cosines, then here is how you find angle X:
29² = 15² + 39² - 2(15)(39)cos(X)
cos(X) = {15² + 39² - 29²}/{2(15)(39)}
X = cos^-1({15² + 39² - 29²}/{2(15)(39)}) = 39.33
For angle Y subtract the other two angles from 180:
Y = 180 - 122 - 39.33 = 18.67
For angle G use the following:
g² = h² + i² - 2(h)(i)cos(G)
cos(G) = {8² + 15² - 13²}/{2(8)(15)}
G = cos^-1({8² + 15² - 13²}/{2(8)(15)}) = 60
For angle H:
H = cos^-1({15² + 13² - 8²}/{2(13)(15)}) = 32.2
For angle I subtract the other two from 180:
I = 180 - 60 - 32.2 = 87.8
z² = 29² + 15² - 2(29)(15)cos(122) = 841 + 225 - (870)cos(122) ≈ 1527
z ≈ 39
I would recommend using the law of sines to find one of the remaining angles but, if you insist on the law of cosines, then here is how you find angle X:
29² = 15² + 39² - 2(15)(39)cos(X)
cos(X) = {15² + 39² - 29²}/{2(15)(39)}
X = cos^-1({15² + 39² - 29²}/{2(15)(39)}) = 39.33
For angle Y subtract the other two angles from 180:
Y = 180 - 122 - 39.33 = 18.67
For angle G use the following:
g² = h² + i² - 2(h)(i)cos(G)
cos(G) = {8² + 15² - 13²}/{2(8)(15)}
G = cos^-1({8² + 15² - 13²}/{2(8)(15)}) = 60
For angle H:
H = cos^-1({15² + 13² - 8²}/{2(13)(15)}) = 32.2
For angle I subtract the other two from 180:
I = 180 - 60 - 32.2 = 87.8