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

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

Sue Leahy
Answered 2021-11-23 Author has 27 answers
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
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