Learning Lab

Examples

1. In triangle \(PQR\) find:

1. the side length \(p\)
   
2. the side length \(q\).
   Solution a.
   Use the sine rule in the form \[\begin{align*} \frac{p}{\sin P} & =\frac{r}{\sin R}\\ \frac{p}{\sin70^{\circ}} & =\frac{15}{\sin30^{\circ}}\\ p & =\frac{15\times\sin70^{\circ}}{\sin30^{\circ}}\\ p & =28.2\;. \end{align*}\] The side length is \(p=28.2\,cm.\)
   Solution b)
   Angle \(Q\) is found using the fact that the sum of the three interior angles of a triangle add to \(180^{\circ}\): \[\begin{align*} \angle Q & =180^{\circ}-\left(70^{\circ}+30^{\circ}\right)\\ & =80^{\circ}. \end{align*}\] From the Sine rule \[\begin{align*} \frac{q}{\sin Q} & =\frac{r}{\sin R}\\ \frac{q}{\sin80^{\circ}} & =\frac{15}{\sin30^{\circ}}\\ q & =\frac{15\times\sin80^{\circ}}{\sin30^{\circ}}\\ q & =29.6\;. \end{align*}\] The length of the side \(q=29.6\,cm.\)

2. In triangle \(ABC\) find:

1. Angle C
   
2. Angle A
   
3. Side length a.
   Solution a.
   Use the sine rule in the form \[\begin{align*} \frac{\sin A}{a} & =\frac{\sin B}{b}=\frac{\sin C}{c} \end{align*}\] The relevant part of the formula is \[\begin{align*} \frac{\sin B}{b} & =\frac{\sin C}{c}\\ \frac{\sin126^{\circ}}{20} & =\frac{\sin C}{12}\\ \frac{\sin126^{\circ}}{20}\times12 & =\sin C\\ \sin C & =0.485\\ C & =\sin^{-1}0.485\\ C & =29^{\circ}. \end{align*}\] Solution b.
   \[\begin{align*} \angle A & =180^{\circ}-\left(126^{\circ}+29^{\circ}\right)\\ & =25^{\circ}. \end{align*}\] Solution c.
   Use the sine rule in the form \[\begin{align*} \frac{a}{\sin A} & =\frac{b}{\sin B}=\frac{c}{\sin C} \end{align*}\] The relevant part of the formula is \[\begin{align*} \frac{a}{\sin A} & =\frac{b}{\sin B}\\ \frac{a}{\sin25^{\circ}} & =\frac{20}{\sin126^{\circ}}\\ a & =\frac{20}{\sin126^{\circ}}\times\sin25^{\circ}\\ a & =10.4m \end{align*}\]

3. Given a a triangle \(ABC\) with angle \(A=30^{\circ}\), side \(c=12\) cm and side \(a=8\) cm, find angle \(C\).
   Solution:
   In this case there are two possible solutions as shown below.

This is called the ambiguous case of the sine rule.
To solve the problem, use the sine rule in the form: \[\begin{align*} \frac{\sin A}{a} & =\frac{\sin C}{c}\\ \frac{\sin30^{\circ}}{8} & =\frac{\sin C}{12}\\ \frac{\sin30^{\circ}}{8}\times12 & =\sin C\\ \sin C & =\frac{1}{16}\times12\\ & =0.75\\ C & =\sin^{-1}\left(0.75\right)\\ & =48.6^{\circ}\textrm{or }131.4^{\circ}. \end{align*}\] If \(C=48.6^{\circ},\) this solution gives the triangle \(ABC_{2}.\)
If \(C=131.4^{\circ}=\left(180^{\circ}-48.6^{\circ}\right)\), this solution gives the triangle \(ABC_{1}.\)

Note: In any non-right-angled triangle, where two sides and the non-included angle are given, check for the ambiguous case.

If the angle is acute and the length of the side adjacent to the angle is greater than the length of the side opposite the angle there may be 2 possible solutions.