Triangles

This section is just an overview of the information that might be learned pre-college. Large books have been dedicated to just the triangle. For more information we could search Clark Kimberling's encyclopedia of triangle centers.

In one way of thinking, all triangles on a plane can be calculated working with right triangles! Look at Fig 1. Any triangle can be rotated until it is possible to construct a rectangle around it, which results in only right angle triangles (such as triangle $A$) outside of the original triangle. If the vertices of the original triangle are known, then the side lengths of the right triangles, $A$, $B$, and $C$ are known by the Pythagorean theorem. The area of the right triangles are $area=base\cdot height/2$. And, of course the rectangle area is known and so by subtraction, the original triangle area is known. Finally, using just the most basic trig definitions for cosine and sine, all of the angles can be computed.

If the vertices of the original triangle were unknown, but the side lengths were known, then one point could be set at the origin and one side could be oriented with an axis. That would lead to two right triangles and again the Pythagorean theorem could be used to solve for the rectangle dimensions. Although in general, the above information, including the two principle identities, $\sin(\theta)=opp/hyp$ and $\cos(\theta)=adj/hyp$ is enough information to solve all three angles and all three sides from any three pieces of information, it is easier if we add the law of cosines, which just generalizes Pythagoras. The law of cosines is not obvious from anything presented here. $$c^{2}=a^{2}+b^{2}-2\cdot a\cdot b\cdot \cos(\theta)$$

In the formula shown, $a$ and $b$ are the sides which include angle $\theta$. Side $c$ is the remaining side. There is no hypotenuse as that is a term that applies only to right triangles. $\theta$ is the angle opposite side $c$.

If only the three angles of a triangle are given, there is no way to know the sides, although if one side is set equal to $x$, then the other two may be designated relative to $x$ through use of the law of sines.

Finally, there exists a law of sines, which from time to time may be convenient. In this formula, it is the side divided by the opposite angle in each term. $${\displaystyle \frac{a}{sin(\alpha)}=\frac{b}{sin(\beta)}=\frac{c}{sin(\theta)}}$$

The circle of radius $r$ is shown in Fig 2. The relationship is that the side along the $x$-axis will have length $r\cdot|\cos(\theta)|$ and the side along the $y$-axis will have length $r\cdot|\sin(\theta)|$. When $r=1$ this becomes the “unit circle”.

Example: Triangle Problem. Given a square of side $a=4$ . Included Triangle $GRM$ is equilateral. What is the area of the shaded triangle in the figure. None of the other information shown in the figure is given by the problem.
Answer: This problem is far more tedious than it is difficult. Indeed that is the case with nearly all triangle solutions. Let's refer to the shaded triangle as $S$.

1. First, we have drawn the red line parallel to the right side to simplify the problem.
2. Note that the right side of shaded area $S$ is $a=4$ and due to the equilateral triangle $(GRM)$, we know the $60^{\circ}$ angle. That gives both of the $30^{\circ}$ angles by subtraction.
3. The line from $G$ to $A$, bisects a right angle so that the lower angle of $S$ is $\pi/4$, as must be its opposite. Thus the angle at $\theta\;(\angle RBA)$ must be $180-30-45=105^{\circ}$
4. Side “b” can be found from the law of sines. $\frac{4}{sin(105)}=\frac{b}{sin(30)}\Longrightarrow b \approx 2.0706$
5. Side $h$ comes from trig. $\sin(45^{\circ})=\frac{h}{b}\Longrightarrow h=1.464$
6. Area is half the base $(4)$ times the height $(h)$. $Area=2\cdot1.464=2.928$

Example: Triangle Problem. This one is on the “more difficult” side. Given triangle $ABD$ and $BCD$. $\angle DBC=\pi/6$ and $\angle ABC=\pi/2$. Length $CD$ is $1$ and length $AB$ is $1$. Find length $AC$.
Answer: There are multiple ways to go about this solution, but here is one. Set point $B$ at the origin, $B=(0,0)$. Then $A$ is on the unit circle and has coordinates $A(x,y)=(\cos(2\pi/3),\sin(2\pi/3))$. Also, point $D$ is partially known, as its $y$ value is $0$. Let $\gamma$ be the distance from point $A$ to point $C$. We will now set up four equations in four unknowns. The unknowns are $C_{x},C_{y},D_{x},\text{ and }\gamma.$ The first equation is the distance formula for $\gamma$. $$\gamma=\sqrt{(A_{x}-C_{x})^{2}+(A_{y}-C_{y})^{2}}$$ The second equation is the line equation for $B$ to $C$. This is known because its slope is essentially given by the angle $\pi/6$ and the intercept is at the origin. $C_{x}$ and $C_{y}$ are the $x,y$ coordinates of point $C$ on the line. $$C_{y}=\frac{\sin\left({\displaystyle \frac{\pi}{6}}\right)}{\cos\left({\displaystyle \frac{\pi}{6}}\right)}\cdot C_{x}$$ For the third equation, the distance from $C$ to $D$ is known. $$1=\sqrt{(C_{x}-D_{x})^{2}+(C_{y}-0)^{2}}$$ For the fourth equation, the distance from $A$ to $D$ has to be $\gamma+1$. $$\gamma+1=\sqrt{(A_{x}-D_{x})+(A_{y}-0)^{2}}$$ Due to the knowing the values for $A_{x}$ and $A_{y}$ these can be simplified to the following set. $$\gamma=\sqrt{{\displaystyle \frac{1}{4}}}\cdot\sqrt{(-2C_{x}-1)^{2}+(-2C_{y}+\sqrt{3})^{2}}$$ $$C_{y}=C_{x}/\sqrt{3}$$ $$1=\sqrt{C_{y}^{2}+(C_{x}-D_{x})^{2}}$$ $$\gamma+1=\sqrt{{\displaystyle \frac{1}{4}}}\cdot\sqrt{(-2D_{x}-1)^{2}+3}$$ I used a CAS to solve the system and got the following results. $$\bbox[yellow]{\gamma=\overline{AC}=\sqrt[3]{2}=1.2599}$$ $$D_{x}=\sqrt[3]{4}=1.5874$$ $$C_{x}=\left(\frac{1}{2}\right)\cdot\left(2^{(2/3)}-2^{(1/3)}+1\right)=0.6637$$ $$C_{y}=\left(\frac{1}{6}\right)\cdot\sqrt{3}\cdot\left(2^{(2/3)}-2^{(1/3)}+1\right)=0.3832$$ Finally, for extra credit, $\overline{BC}=\sqrt{2^{(2/3)}-1}=0.7664$