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21 January, 22:41

In triangle ABC, a = 9, c = 5, and angle B = 120Á. Find the measure of angle A to the nearest degree. Show your work.

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  1. 21 January, 22:51
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    B = sqrt (a^2 + c^2 - 2ac cos B) = sqrt (9^2 + 5^2 - 2 x 9 x 5 x cos 120°) = sqrt (81 + 25 - 90cos120°) = sqrt (106 + 45) = sqrt (151) = 12.29

    b/sin B = a / sin A

    12.29/sin 120° = 9/sin A

    sin A = 9sin 120° / 12.29 = 0.6343

    A = arcsin (0.6343) = 39°
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