# Consider the non-right triangle below 19610800121.jpg Suppose that m\angle ACB=103^{\circ} and m\angle

Consider the non-right triangle below

Suppose that $$\displaystyle{m}\angle{A}{C}{B}={103}^{{\circ}}$$ and $$\displaystyle{m}\angle{B}{A}{C}={44}^{{\circ}}$$, and that $$\displaystyle{y}={50.5}{c}{m}$$. What is the value of x?

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Sue Leahy
Step 1
We know that the sum of angles in a triangle equals $$\displaystyle{180}^{{\circ}}$$ degrees. Therefore,
$$\displaystyle\angle{B}{A}{C}+\angle{C}{B}{A}+\angle{A}{C}{B}={180}^{{\circ}}$$
$$\displaystyle\angle{C}{B}{A}={180}^{{\circ}}-\angle{B}{A}{C}-\angle{A}{C}{B}$$
$$\displaystyle\angle{C}{B}{A}={180}^{{\circ}}-{44}^{{\circ}}-{103}^{{\circ}}$$
$$\displaystyle\angle{C}{B}{A}={33}^{{\circ}}$$
Step 2
By using law of sines in the given triangle,
$$\displaystyle{\frac{{{\sin{\angle}}{B}{A}{C}}}{{{x}}}}={\frac{{{\sin{\angle}}{C}{B}{A}}}{{{y}}}}$$
$$\displaystyle{\frac{{{\sin{{44}}}^{{\circ}}}}{{{x}}}}={\frac{{{\sin{{33}}}^{{\circ}}}}{{{50.5}}}}$$
$$\displaystyle{x}={\frac{{{\sin{{44}}}^{{\circ}}}}{{{\sin{{33}}}^{{\circ}}}}}\times{50.5}$$
$$\displaystyle{x}={\frac{{{0.6946}}}{{{0.5446}}}}\times{50.5}$$
$$\displaystyle{x}={64.41}$$
The value of x is therefore 64.41 cm