Question

Solve the right triangle shown in the image below.

Answer 1

check the picture below.

$\bf tan(A)=\cfrac{2.7}{1.1}\implies \measuredangle A=tan^{-1}\left( \cfrac{2.7}{1.1} \right)\implies \measuredangle A\approx 67.83^o \\\\\\ tan(B)=\cfrac{1.1}{2.7}\implies \measuredangle B=tan^{-1}\left( \cfrac{1.1}{2.7} \right)\implies \measuredangle B\approx 22.17^o \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{2.7^2+1.1^2}\implies c=\sqrt{8.5}\implies c\approx 2.915475947$