Q:

Help find area of parallelogram!!!

Accepted Solution

A:
[tex]\bf \textit{Law of sines} \\\\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{sin(75^o)}{17}=\cfrac{sin(D)}{10}\implies \cfrac{10sin(75^o)}{17}=sin(D) \\\\\\ sin^{-1}\left[ \cfrac{10sin(75^o)}{17} \right]=D\implies 34.6\approx D[/tex]since all interior angles in a triangle must be 180°, that means that C = 180 - 75 - 34.6 = 70.4.  Let's find AD, which is the other sides pair length.[tex]\bf \cfrac{sin(75^o)}{17}=\cfrac{sin(70.4^o)}{AD}\implies ADsin(75^o)=17sin(70.4^o) \\\\\\ AD=\cfrac{17sin(70.4^o)}{sin(75^o)}\implies AD\approx 16.58[/tex]now, check the picture below, let's find the altitude of the parallelogram.[tex]\bf sin(34.6^o)=\cfrac{\stackrel{opposite}{h}}{\stackrel{hypotenuse}{16.58}}\implies 16.58sin(34.6^o)=h\implies 9.4\approx h \\\\[-0.35em] ~\dotfill\\\\ \textit{area of a parallelogram}\\\\ A=bh~~ \begin{cases} b=base\\ h=height\\ \cline{1-1} b=17\\ h=9.4 \end{cases}\implies A=(17)(9.4)\implies A=159.8[/tex]