# Given a triangle with a 112^circ angle. The adjusted sides to this angle are equal. Find their length, if A=80cm^2.

Non-right triangles and trigonometry
Given a triangle with a $$\displaystyle{112}^{\circ}$$ angle. The adjusted sides to this angle are equal. Find their length, if $$\displaystyle{A}={80}{c}{m}^{{2}}$$.

2021-02-04

Let each of the sides be S.
The formula for the triangle with equal sides:
$$\displaystyle{A}=\frac{{1}}{{2}}{S}^{{2}}{{\sin}^{{2}}\theta}$$
Isolate S:
$$\displaystyle{S}=\sqrt{{\frac{{{2}{A}}}{{{{\sin}^{{2}}\theta}}}}}$$
Substitute $$\displaystyle\theta={112}^{\circ}{\quad\text{and}\quad}{A}={80}{c}{m}^{{2}}$$ from the given:
$$\displaystyle{S}=\sqrt{{\frac{{{160}}}{{{{\sin}^{{2}}{\left({112}^{\circ}\right)}}}}}}$$
$$\displaystyle=\frac{{{4}\sqrt{{10}}}}{{\sqrt{{{{\sin}^{{2}}{\left({112}^{\circ}\right)}}}}}}$$
$$\displaystyle=\frac{{{4}\sqrt{{10}}\sqrt{{{{\sin}^{{2}}{\left({112}^{\circ}\right)}}}}}}{{\sqrt{{{{\sin}^{{2}}{\left({112}^{\circ}\right)}}}}\sqrt{{{{\sin}^{{2}}{\left({112}^{\circ}\right)}}}}}}$$
$$\displaystyle=\frac{{{4}\sqrt{{10}}\sqrt{{{{\sin}^{{2}}{\left({112}^{\circ}\right)}}}}}}{{{{\sin}^{{2}}{\left({112}^{\circ}\right)}}}}$$
$$=13.64250...$$