# constructing a spherical triangle using only the laws of sines and cosines

I have a spherical triangle with corners $A,B,C$, angles $\alpha, \beta, \gamma$ and sides $a,b,c$ (which are opposite to the corresponding corners/angles).

I am given $a,b$ (with $a>b$) and $\alpha$. I want to explicitly calculate the other angles and lengths of the triangle by using only the law of sines and the law of cosines for sides and angles.

I am also given a hint, that I should draw a line through point $C$ which intersects side $c$ at right angle (let's call the intersection point $D$). So I want to look at two triangles $A,C,D$ and $B,C,D$ with sides $c', a', b$ and $a, b'', c''$ where $c'+c'' = c$ and $b''=a'$. Their angles are $\alpha, \lrcorner, \gamma'$ and $\beta, \gamma'', \llcorner$ with $\gamma'+\gamma''=\gamma$ and one right angle in each triangle.

My problem is now that especially the law of cosines for sides makes the calculation very complicated and I'm wondering if I'm doing things right.

Starting with the first triangle and the law of sines, I get $$\sin{a'} = \sin{b}\sin{\alpha}\\ \sin{c'} = \sin{b}\sin{\gamma'}$$

Using the law of cosines for angles, I get $3$ reasonable simple expressions but the law of cosines for sides gives me $3$ terms which look like this $$\cos{a'}=\cos{c'}\cos{b} + \sin{c'}\sin{b}\sin{\alpha}$$

I tried to solve the resulting system of $8$ equations but it is really complicated and led me to the nonsensical result that a product of sines has to be zero. This is probably due to an algebraic error of mine. And on top of the algebraic problems, I'll also have to take the inverses of trigonometric functions in the end.

The law of sines yields

$$\sin\beta=\frac{\sin\alpha\sin b}{\sin a}$$

From which you can read $\beta$, or at least two possible values of $\beta$. The cosine law can be applied like this:

\begin{align*} \cos a &= \cos b\cos c + \sin b\sin c\cos\alpha \\ \cos b &= \cos c\cos a + \sin c\sin a\cos\beta \\ \end{align*}

So if you are thinking about solving systems of equations, then solve this:

$$\begin{pmatrix} \cos b & \sin b\cos\alpha \\ \cos a & \sin a\cos\beta \end{pmatrix}\cdot\begin{pmatrix} \cos c \\ \sin c \end{pmatrix} = \begin{pmatrix} \cos a \\ \cos b \end{pmatrix}$$

But what's with that hint? For a right angle, the cosine law simplifies to the spherical version of the pythagorean theorem, and in the law of sines you can drop one denominator, so you have

\begin{align*} \cos b &= \cos a'\cos c' & \frac{\sin c'}{\sin\gamma'} = \frac{\sin a'}{\sin\alpha} &= \sin b \\ \cos a &= \cos a'\cos c'' & \frac{\sin c''}{\sin\gamma''} = \frac{\sin a'}{\sin\beta} &= \sin a \end{align*}

And of course there are the angle versions of the cosine law, which become somewhat simpler as well:

\begin{align*} \cos\alpha &= \sin\gamma'\cos a' & \cos\beta &= \sin\gamma''\cos a' \\ \cos\gamma' &= \sin\alpha\cos c' & \cos\gamma'' &= \sin\beta\cos c'' \end{align*}

So I'd say you can compute $a'$ using either of these:

$$\sin a'=\sin\alpha\sin b=\sin\beta\sin a$$

and once you have that, the other equations will give you everything you need. But again you might have to convert between $\sin$ and $\cos$ so you might have to do case distinctions, and check everything in the end. I'm not sure this is any easier than the version above without the hint.

• Thanks for your answer! So I will try to do the calculation again later on or tomorrow. Just a quick queston: You said that $\beta$ is ambigious and that I might have to do case distinctions. Doesn't the requirement $a>b$ from above make this unnecessary? – Marc Jun 20 '15 at 9:12
• @Marc. It might. The two candidates are $\beta$ and $\pi-\beta$. So when you have $a\le\frac\pi2$ then yes, this is unambigious. However, in a spherical triangle you may well have two angles which are greater than $\frac\pi2$ so unless there is something I overlooked, there may not always be a quick rule to exclude one candidate. – MvG Jun 20 '15 at 9:19