Question

Given triangle ABC, what is the m

Answer 1

Solution

Given the triangle

$\begin{gathered} Using\text{ cosine rule} \\ a^2=b^2+c^2-2bccosA \\ a^2=9^2+11^2-2(9)(11)cos35 \\ a^2=39.08 \\ a=\sqrt{39.08} \\ a=6.25 \\ \end{gathered}$$\begin{gathered} Using\text{ sine rule,} \\ \frac{a}{sinA}=\frac{b}{sinB} \\ \frac{6.25}{sin\text{ 35}}=\frac{9}{sinB} \\ \\ \end{gathered}$$\begin{gathered} sinB=\frac{9\text{ x sin35}}{6.25} \\ \end{gathered}$[tex]\begin{gathered} sinB=0.818195 \\ m