# Consider the non-right triangle below. 19610800111.jpg Suppose that m\angle BCA=70^{\circ}, and that

Consider the non-right triangle below.

Suppose that $$\displaystyle{m}\angle{B}{C}{A}={70}^{{\circ}}$$, and that $$\displaystyle{x}={33}{c}{m}$$ and $$\displaystyle{y}={47}{c}{m}$$ What is the degree measure of $$\displaystyle\angle{A}{B}{C}?$$

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Glenn Cooper
Step 1
Using the cosine law,
$$\displaystyle{A}{B}=\sqrt{{{A}{C}^{{{2}}}+{B}{C}^{{{2}}}-{\left({2}\cdot{A}{C}\cdot{B}{C}{\cos{\angle}}{B}{C}{A}\right)}}}$$
$$\displaystyle{A}{B}=\sqrt{{{47}^{{{2}}}+{33}^{{{2}}}-{\left({2}\cdot{47}\cdot{33}{\cos{{70}}}^{{\circ}}\right)}}}$$
$$\displaystyle{A}{B}=\sqrt{{{2237.05}}}$$
$$\displaystyle{A}{B}={47.29}$$
Step 2
Using the cosine law,
$$\displaystyle\angle{A}{B}{C}={{\cos}^{{-{1}}}{\left({\frac{{{B}{C}^{{{2}}}+{A}{B}^{{{2}}}-{A}{C}^{{{2}}}}}{{{2}\times{B}{C}\times{A}{B}}}}\right)}}$$
$$\displaystyle\angle{A}{B}{C}={{\cos}^{{-{1}}}{\left({0.3576}\right)}}$$
$$\displaystyle\angle{A}{B}{C}={69.04}^{{\circ}}$$