Question

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.

Answer 1

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