Given a non-right triangle with vertex C = 91 &#x2218;<!-- ∘ --> </msup> , <

Manteo2h 2022-06-26 Answered
Given a non-right triangle with vertex C = 91 ,   A C ¯ = 39.9   c m ,   B C ¯ = 32.6   c m, find
a) A B ¯
b) m A
c) m B
I 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?
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Answers (2)

knolsaadme
Answered 2022-06-27 Author has 16 answers
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
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xonycutieoxl1
Answered 2022-06-28 Author has 7 answers
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.
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