Question

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If triangle ABC has A = 113, side a = 200, and side b = 134, then which of the following is true?


A. angle B=38.08 degrees 
B. angle B=141.92 degrees
C. angle B=38.08 degrees or 141.92 degrees
D. no solution 

Answer 1

For this case we will define the following:
 A, B: are the values of the triangle angles
 a, b: are the values of the lengths of the sides opposite the angles A, B.
 We then have to use the law of the sine:
 $\frac{sinB}{b} = \frac{sinA}{a}$
 Clearing the angle B we have:
 $sinB = \frac{sinA}{a}*b$
 $B = arcsin(\frac{sinA}{a}*b)$
 Substituting values we have:
 $B = arcsin(\frac{sin113}{200}*134)$
 $B = 38.08$
 Answer:
 the following is true:
 A. angle B=38.08 degrees 

Answer 2

Answer:

angle B=38.08 degrees

Step-by-step explanation:

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