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Trig. Use law of cosines in degree mode.

First find alpha; the angle opposite a

a^2 = b^2 + c^2 - 2bc*cos(alpha)

24^2 = 36^2 + 19^2 - 2(36)(19)cos(alpha)

576 = 1657 - 1368cos(alpha)

subtract 1657 from both sides( order of operations )

-1088 = -1368cos(alpha)

0.7902046784 = cos(alpha)

arccos(0.7902046784) = alpha

38 degrees = alpha ( angle opposite side a )

find beta; angle opposite side b

b^2 = a^2 + c^2 - 2ac*cos(beta)

1296 = 937 - 912cos(beta)

359 = -912cos(beta)

-0.3936403509 = cos(beta)

arcos(-0.3936403509 = beta

113 degrees = beta ( angle opposite of b )

easy thing to get last angle

180 degrees - 38 degrees - 113 degrees

= 29 degrees; which is gamma; angle opposite c

alpha( angle opposite a side = 38 degrees

beta( angle opposite b side ) = 113 degrees

gamma(angle opposite c side) = 29 degrees

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Q: Find the missing angles for triangle ABC given that a equals 24 b equals 36 and c equals 19?
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