Finding all sides and angles of a triangle

Question

So SAS, SSS, ASA, AAS and RHS are reasons for congruent triangles, that means if a triangle, for example, have side lengths of 5, 6 and 8, then the triangle is unique. What I am trying to do is to find an expressions for other sides and angles in terms of the given sides and angles. I've solved it but some of the expressions are a bit long and ugly, so can anyone verify my workings and simplify them if possible? Any help would be appreciated :)

For triangles below, greek letters are angles and english letters are sides

$\alpha$ is the opposite angle of side a

$\beta$ is the opposite angle of side b

$\gamma$ is the opposite angle of side c

SAS: given sides $a$, $b$ and included angle $\gamma$

$$c^2=a^2+b^2-2ab\cos{(\gamma)}$$ $$\boxed{c=\sqrt{a^2+b^2-2ab\cos{(\gamma)}}}$$ $$\frac{\sin{(\alpha)}}{a}=\frac{\sin{(\gamma)}}{c}$$ $$\frac{\sin{(\alpha)}}{a}=\frac{\sin{(\gamma)}}{\sqrt{a^2+b^2-2ab\cos{(\gamma)}}}$$ $$\sin{(\alpha)}=\frac{a\sin{(\gamma)}}{\sqrt{a^2+b^2-2ab\cos{(\gamma)}}}$$ $$\boxed{\alpha=\arcsin{\left(\frac{a\sin{(\gamma)}}{\sqrt{a^2+b^2-2ab\cos{(\gamma)}}}\right)}}$$ $$\frac{\sin{(\beta)}}{b}=\frac{\sin{(\gamma)}}{c}$$ $$\frac{\sin{(\beta)}}{b}=\frac{\sin{(\gamma)}}{\sqrt{a^2+b^2-2ab\cos{(\gamma)}}}$$ $$\boxed{\beta=\arcsin{\left(\frac{b\sin{(\gamma)}}{\sqrt{a^2+b^2-2ab\cos{(\gamma)}}}\right)}}$$ SSS: given sides $a$, $b$ and $c$ $$c^2=a^2+b^2-2ab\cos{(\gamma)}$$ $$a^2+b^2-c^2=2ab\cos{(\gamma)}$$ $$\boxed{\gamma=\arccos{\left(\frac{a^2+b^2-c^2}{2ab}\right)}}$$ $$b^2=a^2+c^2-2ac\cos{(\beta)}$$ $$a^2+c^2-b^2=2ac\cos{(\beta)}$$ $$\boxed{\beta=\arccos{\left(\frac{a^2+c^2-b^2}{2ac}\right)}}$$ $$a^2=b^2+c^2-2bc\cos{(\alpha)}$$ $$b^2+c^2-a^2=2bc\cos{(\alpha)}$$ $$\boxed{\alpha=\arccos{\left(\frac{b^2+c^2-a^2}{2bc}\right)}}$$ ASA: given included side $c$ and angles $\alpha$, $\beta$ $$\alpha+\beta+\gamma=\pi$$ $$\boxed{\gamma=\pi-\alpha-\beta}$$ $$\frac{\sin{(\alpha)}}{a}=\frac{\sin{(\gamma)}}{c}$$ $$\frac{\sin{(\alpha)}}{a}=\frac{\sin{(\pi-\alpha-\beta)}}{c}$$ $$\boxed{a=\frac{c\sin{(\alpha)}}{\sin{(\alpha+\beta)}}}$$ $$\frac{\sin{(\beta)}}{b}=\frac{\sin{(\gamma)}}{c}$$ $$\frac{\sin{(\beta)}}{b}=\frac{\sin{(\pi-\alpha-\beta)}}{c}$$ $$\boxed{b=\frac{c\sin{(\beta)}}{\sin{(\alpha+\beta)}}}$$ AAS: given non-included side $a$ and angles $\alpha$, $\beta$ $$\alpha+\beta+\gamma=\pi$$ $$\boxed{\gamma=\pi-\alpha-\beta}$$ $$\frac{\sin{(\beta)}}{b}=\frac{\sin{(\alpha)}}{a}$$ $$\boxed{b=\frac{a\sin{(\beta)}}{\sin{(\alpha)}}}$$ $$\frac{\sin{(\gamma)}}{c}=\frac{\sin{(\alpha)}}{a}$$ $$\boxed{c=\frac{a\sin{(\gamma)}}{\sin{(\alpha)}}}$$ RHS: given shorter side $a$, hypotenuse $c$ and right angle $\gamma=\frac{\pi}{2}$ $$\sin{(\alpha)}=\frac{a}{c}$$ $$\boxed{\alpha=\arcsin{\left(\frac{a}{c}\right)}}$$ $$\alpha+\beta+\gamma=\pi$$ $$\arcsin{\left(\frac{a}{c}\right)}+\beta+\frac{\pi}{2}=\pi$$ $$\boxed{\beta=\arccos{\left(\frac{a}{c}\right)}}$$ $$a^2+b^2=c^2$$ $$\boxed{b=\sqrt{c^2-a^2}}$$ For ASA, AAS and RHS, it isn't really that ugly but I typed them out anyways...

Edit: for $\arcsin$, the angle could be the supplement of the result depending on the angle is acute or obtuse.

Answer 1

To say an alternative method for SSS, you can find the area with Heron's formula:

$$A=\sqrt{s(s-a)(s-b)(s-c)},$$ where $s=\dfrac{a+b+c}2$.

Then you can find the angles this way (assume that $a\le b\le c$): $$A=\frac{bc\sin\alpha}2\implies \alpha=\arcsin\frac{2A}{bc}$$ $$A=\frac{ac\sin\beta}2\implies \beta=\arcsin\frac{2A}{ac}$$ $$\gamma=\pi-\alpha-\beta$$

In addition, you can find circunradius $R$ and inradius $r$: $$R=\frac{b}{2\sin\beta}$$ (you can use any pair of side and opposite angle here: choose the angle nearest to $\pi/2$ for better precission: it should be $\beta$ or $\gamma$). $$r=\frac{2A}{a+b+c}$$

Furthermore, $$\frac Rr=\frac{c(a+b+c)}{4A\sin\gamma}$$ But the altitude $h_c$ is the shortest altitude of the triangle, and hece is shorter than every side. And $0<\sin\gamma\le 1$, so $$\frac Rr=\frac{c(a+b+c)}{4\dfrac{ch_c}2\sin\gamma}=\frac{a+b+c}{2h_c\sin\gamma}>\frac32$$

I think that this bound can be improved but I don't know how.

Answer 2

For the SAS case your formulas are going to get even uglier when you take into account all the possibilities of obtuse angles. But you can simplify things somewhat by re-using $c$ after you compute it: \begin{align} c &= \sqrt{a^2+b^2-2ab\cos{(\gamma)}}, \\ \alpha &= \begin{cases} \arcsin\left(\dfrac{a\sin(\gamma)}{c}\right) & \text{if $a\cos(\gamma)<b$},\\ \pi - \arcsin\left(\dfrac{a\sin(\gamma)}{c}\right) & \text{otherwise},\\ \end{cases}\\ \beta &= \pi - \alpha - \gamma. \end{align}

You could reduce the formula for $\alpha$ to one case by using the law of cosines again, but a simpler way is to express the problem like this:

Given sides $a \leq b$ and included angle $\gamma$, \begin{align} c &= \sqrt{a^2+b^2-2ab\cos{(\gamma)}}, \\ \alpha &= \arcsin\left(\dfrac{a\sin(\gamma)}{c}\right),\\ \beta &= \pi - \alpha - \gamma. \end{align}

This works because $a < b$ implies $a\cos(\gamma)<b$.

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For RHS, knowing that $a$ is the shorter side, you can compute $$ \alpha=\arcsin{\left(\frac{a}{c}\right)}, \\ \beta = \frac\pi2 - \alpha, \\ b=c\cos{(\alpha)}. $$ This is a more accurate computation of $\beta$ in general, as the arc cosine for numbers near $1$ can lose precision.

Note that your formula for $b$ uses a previously-computed value $\alpha$ rather than only the input data. That's not a criticism; it's simply a note that you already implicitly understand the benefit of re-using computations rather than fully expanding every formula in terms of the initially given variables.

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My advice for the SSS case is to follow the previous answer to your question. I would argue that this method does give you the answers in terms of the three sides, because ultimately everything in the method is computed from $a$, $b$, and $c$ with no other information from outside the calculations. Yes, it introduces additional symbols, but they are merely intermediate results of the formulation.

Of course by substituting $\dfrac{a+b+c}2$ for $s$ in the equation for $A$ in Heron's formula, then substituting that result for $A$ in the equations for $\alpha$ and $\beta$, and simplifying, you could write $$ \alpha=\arcsin\frac{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}}{2bc}, \\ \beta=\arcsin\frac{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}}{2ac}, $$ which is fewer equations, but it is more symbols and operations (count them) and you have to carefully check that it really is the exact same expression $\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}$ in both formulas if you want to save the effort of computing it twice. You also lose sight of the notions of semiperimeter and area on which the formula was really based.

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For ASA and AAS you can identify shared parts of the equations for the unknown sides that can be given a name in a separate equation, but this does not gain you much. For ASA you might write $\sin(\gamma)$ instead of $\sin(\alpha + \beta)$.