# 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?

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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