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