Solution of triangles

We talk about solution of triangles when we want to solve a triangle, in other words, when we want to find out all sides and angles of a triangle. To do this we will use:

  1. The law of sines.
  2. The law of cosines.
  3. The fact that the sum of the angles of a triangle is always $$180$$ degrees (or $$\pi$$ rad).

Now, we are going to classify all the possible cases that we may find when trying to solve the triangle. Note, however, that thanks to point $$(3)$$, once we know two of the angles we can easily deduce the third one. Thus, we will say that a triangle is solved if we can determine two of its angles and the length of each of the three sides.

  1. Given a triangle of which we know $$3$$ sides and:
    1. $$1$$ angle: From the law of sines or that of cosines we calculate another angle and, consequently, we have solved the triangle.
    2. $$0$$ angles: From the law of cosines, we can calculate one angle. Then we are in the case $$1.1.$$
  2. Given a triangle of which we know $$2$$ sides and:
    1. $$2$$ angles: From the law of cosines or sines we find the third side, solving the triangle.
    2. $$1$$ angle: It depends on which is the angle we know.
    3. If the known angle is the one formed by the two known sides, we will apply the law of cosines, obtaining the third side and we will proceed as in the case $$1.1$$.
    4. If the known angle is not the one formed by the two known sides, we will have two options: using the law of cosines and solving the quadratic equation to obtain the third side and then proceed as in $$(1.i)$$, or using the law of sines to find another angle and proceed as in $$(2.1)$$. It is necessary to emphasize that in this case the solution could not be unique. By means of the law of cosines, we will have to solve a quadratic equation and, therefore, we can have two positive solutions (with geometric sense). On the other hand, whenever we use the law of sines to find an angle we also obtain two possible angles, since the sine has two solutions that are in the 1st and 2nd quadrant, which, therefore, are valid since these two angles are less than $$180$$ degrees (something that does not happen with the cosine, since the other solution belongs to the 4th quadrant and does not make sense geometrically). We can say as a conclusion that, in this case, we can find $$2$$, or $$1$$, or no solution(s).
    5. $$0$$ angles: Infinite solutions exist, we will not be able to solve the triangle.
  3. Given a triangle of which we know $$1$$ side and:
    1. $$2$$ angles: By means of the law of sines we find another side and proceed like $$(2.1)$$.
    2. $$1$$ or $$0$$ angles: There are infinite solutions; we will not be able to solve the triangle.
  4. Given a triangle about which we know $$0$$ sides and:
    1. $$2$$ angles: There are infinite solutions, that is, there exist infinite triangles that have these two angles, but they all are similar.
    2. $$1$$ or $$0$$ angles: There are infinite solutions, we will not be able to solve the triangle.

We are going to solve a case in which we know one of the sides, $$a=5$$ cm, and two angles $$B=45^\circ$$ and $$C=60^\circ$$. Therefore we are in case $$3.1.$$

So, we apply the law of sines, but for that we need to know the third angle, but, as we have already said, knowing two angles is almost the same as knowing all the angles since $$A+B+C=180^\circ$$ and therefore: $$A=180^\circ-60^\circ-45^\circ=75^\circ$$.

We have: $$$\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B} \Rightarrow b=\frac{5 \sin 45^\circ}{sin 74^\circ}=\frac{5 \cdot \frac{\sqrt{2}}{2}}{0.9656}=3.66 \mbox{ cm }$$$ Now we already have $$2$$ angles and two sides, therefore, we are in point $$2.1$$ and consequently we apply the law of cosines to find the third side: $$$\displaystyle c^2=a^2+b^2-2bc\cos C=25+13.40-36.60 \cdot \cos 60=$$$ $$$\displaystyle =25+13.40-36.60\cdot \frac{1}{2}=20.10 \Rightarrow c= 4.48 \mbox{ cm }$$$ Thus, we have solved the triangle.