PLEASE HELP, ILL MARK BRAINLIEST ""solve triangle ABC using the given information round triangle measures to the nearest degree and side measure to the nearest tenth ​

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  a) A = 73°, C = 59°, a = 16.7

  b) A = 11°, C = 121°, a = 3.3

Step-by-step explanation:

The Law of Sines can be used to find angle C. Then the sum of angles in the triangle can be used to find angle A. The length of 'a' can be found either from the Law of Sines or the Law of Cosines at that point.

The given angle is opposite the shorter of the two given sides, so there are two possible solutions.

Acute triangle solution

The Law of Sines gives the relation ...

  sin(C)/c = sin(B)/b

  sin(C) = (c/b)sin(B)

  C = arcsin(c/b·sin(B)) = arcsin(15/13·sin(48°)) ≈ 59°

Then angle A is 132° -59° = 73°, and side 'a' is ...

  a = 13·sin(A)/sin(B) = 13·sin(73°)/sin(48°) ≈ 16.7

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Obtuse triangle solution

The other value of arcsin(15/13·sin(48°)) is used:

  C ≈ 121°   ⇒   A = 11°

  a = 13·sin(11°)/sin(48°) ≈ 3.3

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The two solutions are ...

  A = 73°, C = 59°, a = 16.7

  A = 11°, C = 121°, a = 3.3

We note that the drawing is of an acute triangle, so that solution may be the one that is preferred.

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Additional comment

In the above presentation, we have rounded angle values to the precision required by the problem statement (nearest degree). However, in the actual calculations, we have used the full calculator precision. Only the final answers have been rounded.

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step-by-step explanation: