Consider the non-triangle below. Suppose that /_ACB=103^@ and m /_BAC=44^@, and that y=50.5cm. What of the value of X? 01510101661.png

Question
Consider the non-triangle below. Suppose that \(\displaystyle\angle{A}{C}{B}={103}^{\circ}\) and \(\displaystyle{m}\angle{B}{A}{C}={44}^{\circ}\), and that y=50.5cm. What of the value of X?
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Answers (1)

2021-01-17
We know that the sum of angles in a triangle equals 180 degrees. Therefore,
\(\displaystyle\angle{B}{A}{C}+\angle{c}{b}{a}+\angle{A}{C}{B}={180}^{\circ}\)
\(\displaystyle\angle{C}{B}{A}={18}^{\circ}-\angle{B}{A}{C}-\angle{A}{C}{B}\)
\(\displaystyle\angle{C}{B}{A}={180}^{\circ}-{44}^{\circ}-{103}^{\circ}\)
\(\displaystyle\angle{C}{B}{A}={33}^{\circ}\)
By using law of sines in the given triangle,
\(\displaystyle{\left(\frac{{{\sin}\angle{B}{A}{C}}}{{x}}\right)}={\left(\frac{{{\sin}\angle{C}{B}{A}}}{{y}}\right)}\)
\(\displaystyle{\left(\frac{{{\sin{{44}}}^{\circ}}}{{x}}\right)}={\left(\frac{{{\sin{{33}}}^{\circ}}}{{50.5}}\right)}\)
\(\displaystyle{x}=\frac{{{\sin{{44}}}^{\circ}}}{{{\sin{{33}}}^{\circ}}}\cdot{50.5}\)
\(\displaystyle{x}=\frac{{{0.6946}}}{{{0.5446}}}\cdot{50.5}\)
\(\displaystyle{x}={64.41}\)
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