# Consider the non-right triangle below 19610800131.jpg Suppose that m\angle ACB=98^{\circ} and m\angle

Consider the non-right triangle below

Suppose that $$\displaystyle{m}\angle{A}{C}{B}={98}^{{\circ}}$$ and $$\displaystyle{m}\angle{B}{A}{C}={42}^{{\circ}}$$, and that $$\displaystyle{y}={51.4}{c}{m}$$. What is the value of x?

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Mary Darby
Step 1
Solution: in $$\displaystyle\triangle{A}{B}{C}$$ we know that
$$\displaystyle\angle{A}+\angle{B}+\angle{C}={180}^{{\circ}}$$
Now in $$\displaystyle\triangle{A}{B}{C}$$ (given)
$$\displaystyle\angle{B}{A}{C}+\angle{A}{C}{B}+\angle{A}{B}{C}={180}^{{\circ}}$$
$$\displaystyle{42}^{{\circ}}+{98}^{{\circ}}+\angle{A}{B}{C}={180}^{{\circ}}$$
$$\displaystyle\angle{A}{B}{C}={180}^{{\circ}}-{140}^{{\circ}}$$
$$\displaystyle\angle{A}{B}{C}={40}^{{\circ}}$$
Using Sine angle rule
$$\displaystyle{\frac{{{a}}}{{{\sin{{A}}}}}}={\frac{{{b}}}{{{\sin{{B}}}}}}={\frac{{{c}}}{{{\sin{{c}}}}}}$$
Step 2
$$\displaystyle{\frac{{{x}}}{{{\sin{\angle}}{B}{A}{C}}}}={\frac{{{y}}}{{{\sin{\angle}}{A}{B}{C}}}}={\frac{{{A}{B}}}{{{\sin{\angle}}{A}{C}{B}}}}$$
$$\displaystyle{\frac{{{x}}}{{{\sin{{42}}}^{{\circ}}}}}={\frac{{{51.4}}}{{{\sin{{40}}}^{{\circ}}}}}={\frac{{{A}{B}}}{{{\sin{{98}}}^{{\circ}}}}}$$
taking first turo ratio
$$\displaystyle{\frac{{{x}}}{{{0.67}}}}={\frac{{{5.4}}}{{{0.64}}}}$$
$$\displaystyle\Rightarrow{x}={\frac{{{51.4}\times{0.67}}}{{{0.64}}}}$$
$$\displaystyle\Rightarrow{x}={53.73}{c}{m}$$