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Two angles of a triangle measure 30 and 45 degrees. If the side of the triangle opposite the 30-degree angle measures $6\sqrt2$ units, what is the sum of the lengths of the two remaining sides? Express your answer as a decimal to the nearest tenth.

 Mar 9, 2021
 #1
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Two angles of a triangle measure 30 and 45 degrees. If the side of the triangle opposite the 30-degree angle measures 6√2 units, what is the sum of the lengths of the two remaining sides? Express your answer as a decimal to the nearest tenth.

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∠A = 30º        ∠B = 45º         ∠C = 180 - 30 - 45 = 105º

 

a = 6√2

 

a/b = sin∠A / sin∠B             b = 6√2 * sin45º / sin30º         b = 12

 

a/c = sin∠A / sin∠C             c = 6√2 * sin105º / sin 30º       c ≈ 16.4

 

b + c ≈ 28.4          (This answer is ≈ correctsmiley)

 

Pleeeeease, don't post this question again. laugh

 Mar 10, 2021
edited by jugoslav  Mar 10, 2021

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