Given a non-right triangle with vertex C = 91 ∘ , A C ¯ = 39.9 c m , B C ¯ = 32.6 c m, finda) A B ¯ b) m ∡ Ac) m ∡ BI was able to find the length of A B ¯ , which I labeled c c = 32.6 2 + 39.9 2 − 2 ( 32.6 ∗ 39.9 ) cos ⁡ ( 91 ∘ ) c = 51.9 c m but I'm having a hard time finding the angle measures of A and B, here's my attempt:a) 32.6 sin ⁡ A = 51.9 sin ⁡ 91 ∘ sin ⁡ A = 32.6 sin ⁡ 91 ∘ 51.9 So given some theta A, we get the vertical length 0.63 radians. To find A, I used arcsin ⁡ ( 0.63 ) ≈ 0.68 rad So I figured I just needed to convert this to degree measure, getting me m ∡ A ≈ 38.96 ∘ b) using the fact that the sum of angle measures of a triangle is 180 ∘ , I tried to solve for angle B using 38.96 ∘ + 51.9 ∘ + m ∡ B ≈ 180 ∘ m ∡ B ≈ 89.14 ∘ but I'm getting marked wrong for both answers b) and c). What am I doing wrong?

Question

Given a non-right triangle with vertex   C  =      91      ∘    ,               A      C        ¯    =  39.9     c  m  ,               B      C        ¯    =  32.6     c  m, finda)             A      B        ¯  b)   m  ∡  Ac)   m  ∡  BI was able to find the length of             A      B        ¯  ,  which I labeled   c                    c        =                                                   32.6            2                    +                      39.9            2                    −          2          (          32.6          ∗          39.9          )          cos          ⁡          (                      91            ∘                    )                                    c        =                             51.9                 c        m            but I&#039;m having a hard time finding the angle measures of   A and   B, here&#039;s my attempt:a)                               32.6                      sin            ⁡            A                          =                              51.9                      sin            ⁡                          91              ∘                                                                        sin        ⁡        A        =                                          32.6            sin            ⁡                          91              ∘                                51.9                    So given some theta A, we get the vertical length 0.63 radians. To find A, I used                    arcsin        ⁡        (        0.63        )        ≈        0.68                 rad            So I figured I just needed to convert this to degree measure, getting me  m  ∡  A  ≈      38.96    ∘  b) using the fact that the sum of angle measures of a triangle is       180    ∘  , I tried to solve for angle   B using                              38.96          ∘                +                  51.9          ∘                +        m        ∡        B        ≈                                       180          ∘                                    m        ∡        B        ≈                                       89.14          ∘                    but I&#039;m getting marked wrong for both answers b) and c). What am I doing wrong?

knolsaadme · Accepted Answer

Side length   c  ≈  51.9 is ok.The triangle side lengths are close to the right triangle of sides 30,40,50 in proportion.  ∠  A  (  ≈      38.96      ∘    )  +  ∠      B      ∘    +      91      ∘    =      180      ∘    →  ∠  B  ≈      50.04      ∘

xonycutieoxl1 · Accepted Answer

On the last answer you are using 51.9, which is the length of side c, instead of the correct value of 91 degrees. As for calculating the angle A, I am guessing that it is marked wrong because of an insufficient explanation. You indeed show that   s  i  n  (  A  )  =  0.63, but this does not always imply that   A  =  38.96 degrees.

Given a non-right triangle with vertex C = 91 ∘ </msup> , <

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