Solution

Use the sine rule $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ to find $B$.

Substitute the given values of $a$, $A$ and $b$. \[\frac{3}{\sin 40^{\circ} } = \frac{4}{\sin B}\] Multiply both sides by $\sin B$ and $\sin 40^{\circ}$. \begin{align} 3\sin B &= 4 \sin 40^{\circ}\\\\ \sin B &= \frac{4 \sin 40^{\circ} }{3}\\ B &= \sin^{-1} \left(\frac{4 \sin 40^{\circ} }{3}\right)\\ B &= 58.99^{\circ} \text{ (to 2 d.p.)} \end{align}

Solution

First, use $\dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ to find $c$.

\begin{align} \frac{6}{\sin 77^{\circ} } &= \frac{c}{\sin 34^{\circ} }\\\\ \frac{6 \sin 34^{\circ} }{\sin 77^{\circ} } &= c\\ c &= 3.44 \text{ (to 3 sig.fig.)} \end{align}

To find $A$, use the fact that the angles in a triangle add up to $180^{\circ}$, i.e. $A+B+C = 180$.

\begin{align} 180 &= A +77 + 34\\ A &= 180 - 77- 34\\ A &= 69^{\circ} \end{align}

Now, use this value for $A$ in the sine rule to find $a$.

\begin{align} \frac{a}{\sin 69^{\circ} } &= \frac{6}{\sin 77^{\circ} }\\\\ a &= \frac{6 \sin 69^{\circ} }{\sin 77^{\circ} }\\\\ a &= 5.75 \text{ (to 3 sig.fig.)} \end{align}

Solution

Use $a^2 = b^2 + c^2 -2bc\cos A$.

\begin{align} a^2 &= 2^2 + 5^2 - (2\times2\times5)\cos 39^{\circ}\\ a^2 &= 29 - 20\cos 39^{\circ}\\ a^2 &= 13.457 \text{ (to 3 d.p.)}\\ a &= 3.67 \text{ (to 3 sig.fig.)} \end{align}

Solution

The angle between $a$ and $b$ is labelled $C$ in the diagram above, so use $c^2 = a^2 + b^2 - 2ab\cos C$.

\begin{align} 9^2 &= 7^2 + 6^2 - (2\times7\times6)\cos C\\ 81 &= 49+36 - 84\cos C\\ 84\cos C &= 4\\ \cos C &= \frac{4}{84} = \frac{1}{21}\\\\ C &= \cos^{-1}\left(\frac{1}{21}\right)\\ C &= 87.3^{\circ} \text{ (to 3 sig.fig.)} \end{align}