Question 93616
If a given triangle ABC with a = 12, b = 10 and c = 8, please help me find the angles in minutes.
:
Using the law of cosines: a^2 = b^2 + c^2 - 2bc*Cos(A)
:
Rearrange formula to find Angle A
Cos(A) = {{{(b^2 + c^2 - a^2)/(2bc)}}}
:
Cos(A) = {{{(10^2 + 8^2 - 12^2)/(2*10*8)}}}
:
Cos(A) = {{{(100 + 64 - 144)/(160)}}} = {{{20/160}}} = .125
:
A = 82.81924422 degrees
Find the minutes, mult the decimal portion times 60:
A = 82 degree, 49.15 minutes
:
:
Find Angle B
Cos(B) = {{{(a^2 + c^2 - b^2)/(2ca)}}}
:
Cos(B) = {{{(12^2 + 8^2 - 10^2)/(2*8*12)}}}
:
Cos(B) = {{{(144 + 64 - 100)/(192)}}} = {{{108/192}}} = .5625
:
B = 55.77113367 degrees
Find the minutes, mult the decimal portion times 60:
B = 55 degrees, 46.28 minutes
:
:
Find Angle C by subtracting the known angles from 180:
:
180 - 82.81924422 - 55.77113367 = 41.40962211
Which is:
C = 41 degrees, 24.58 minutes
:
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