Using the cosines law
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Using the cosines law

[From: ] [author: ] [Date: 12-08-27] [Hit: ]
X = cos^-1({15² + 39² - 29²}/{2(15)(39)}) = 39.Y = 180 - 122 - 39.33 = 18.H = cos^-1({15² + 13² - 8²}/{2(13)(15)}) = 32.I = 180 - 60 - 32.2 = 87.......
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 :/

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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
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