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The area of a triangle is 2. Two of the side lenghts are 8.4 and 1.3 and the included angle is acute. Find the measure of the included angle, to the nearest tenth of a degree.

Sagot :

[tex]\begin{gathered} area=\text{ 2} \\ side\text{ 1 = 8.4} \\ \text{side 2 =1.3} \\ \text{triangles area =}\frac{high\cdot base}{2}\text{, the base is 8.4} \\ \text{2}\cdot\text{ triangles area=high}\cdot base \\ \rbrack\text{high}=\frac{\text{2}\cdot\text{ triangles area}}{base} \\ \text{high}=\frac{2\cdot2}{8.4} \\ \text{high}=\text{ 0}.5 \\ sin(\theta)=\frac{opposite}{hypotenuse} \\ In\text{ this case opposite is the high of the triangle, and hypotenuse is 1.3} \\ sin(\theta)=\frac{0.5}{1.3} \\ \theta=\sin ^{-1}(\frac{0.5}{1.3}) \\ \theta=22.6 \\ \text{The angle is 22.6\degree} \end{gathered}[/tex]

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