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

danrussekme 2021-11-22 Answered
Consider the non-right triangle below
image
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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Expert Answer

Mary Darby
Answered 2021-11-23 Author has 1852 answers
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
\(\displaystyle{\frac{{{x}}}{{{0.67}}}}={\frac{{{5.4}}}{{{0.64}}}}\)
\(\displaystyle\Rightarrow{x}={\frac{{{51.4}\times{0.67}}}{{{0.64}}}}\)
\(\displaystyle\Rightarrow{x}={53.73}{c}{m}\)
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