Question

How would I find a given triangle where: A=0.1, B=1 and c=9? Thanks in advance.

Answer 1

$\bf \textit{Law of sines} \\ \quad \\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\\\\ -------------------------------\\\\ \begin{cases} \measuredangle A=0.1\\ \measuredangle B=1 \end{cases}\quad thus\implies \measuredangle C=\pi -A-B\implies \measuredangle C=\pi -1.1$

$\bf \cfrac{sin(B)}{b}=\cfrac{sin(C)}{c}\implies \cfrac{sin(1)}{b}=\cfrac{sin(\pi -1.1)}{9} \\\\\\ \boxed{\cfrac{9sin(1)}{sin(\pi -1.1)}=b} \\\\\\ \cfrac{sin(A)}{a}=\cfrac{sin(C)}{c}\implies \cfrac{sin(0.1)}{a}=\cfrac{sin(\pi -1.1)}{9} \\\\\\ \boxed{\cfrac{9sin(0.1)}{sin(\pi -1.1)}=a}$

make sure your calculator is in Radian mode.