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 $m\mathrm{\angle }ACB={98}^{\circ }$ and $m\mathrm{\angle }BAC={42}^{\circ }$, and that $y=51.4cm$. What is the value of x?
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Step 1
Solution: in $\mathrm{△}ABC$ we know that
$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$
Now in $\mathrm{△}ABC$ (given)
$\mathrm{\angle }BAC+\mathrm{\angle }ACB+\mathrm{\angle }ABC={180}^{\circ }$
${42}^{\circ }+{98}^{\circ }+\mathrm{\angle }ABC={180}^{\circ }$
$\mathrm{\angle }ABC={180}^{\circ }-{140}^{\circ }$
$\mathrm{\angle }ABC={40}^{\circ }$
Using Sine angle rule
$\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}c}$
Step 2
$\frac{x}{\mathrm{sin}\mathrm{\angle }BAC}=\frac{y}{\mathrm{sin}\mathrm{\angle }ABC}=\frac{AB}{\mathrm{sin}\mathrm{\angle }ACB}$
$\frac{x}{{\mathrm{sin}42}^{\circ }}=\frac{51.4}{{\mathrm{sin}40}^{\circ }}=\frac{AB}{{\mathrm{sin}98}^{\circ }}$
taking first turo ratio
$\frac{x}{0.67}=\frac{5.4}{0.64}$
$⇒x=\frac{51.4×0.67}{0.64}$
$⇒x=53.73cm$