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?

Suppose that

danrussekme
2021-11-22
Answered

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?

Suppose that

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Mary Darby

Answered 2021-11-23
Author has **11** answers

Step 1

Solution: in$\mathrm{\u25b3}ABC$ we know that

$\mathrm{\angle}A+\mathrm{\angle}B+\mathrm{\angle}C={180}^{\circ}$

Now in$\mathrm{\u25b3}ABC$ (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

$\frac{x}{0.67}=\frac{5.4}{0.64}$

$\Rightarrow x=\frac{51.4\times 0.67}{0.64}$

$\Rightarrow x=53.73cm$

Solution: in

Now in

Using Sine angle rule

Step 2

taking first turo ratio

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