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Use the law of sines to solve the triangle if possible

Use The Law Of Sines To Solve The Triangle If Possible class=

Sagot :

The Sine rule formula is,

[tex]\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}[/tex]

Given:

[tex]\begin{gathered} a=13.9in \\ A=83^0 \\ b=18.3in \\ B=? \end{gathered}[/tex]

Let us solve for solve for the measure of angle B

[tex]\frac{13.9}{sin83^0}=\frac{18.3}{sinB}[/tex]

Therefore,

[tex]\begin{gathered} sinB=\frac{18.3\times sin83^0}{13.9}=1.30673342266 \\ \therefore B=sin^{-1}(1.30673342266)=No\text{ solution} \end{gathered}[/tex]

Hence, there are no possible solutions for this triangle (OPTION C).

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