Extending to Non-right Triangles

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A pair of examples extending right triangle tools to non-right triangles.

We will now extend our techniques for understanding right triangles to understand non-right triangles. The following two problems will provide a walk through for how to find right triangles that we can use in a picture of an arbitrary triangle.

1 : We will first find the missing sides and angles of the triangle below given that $A=\sage {A1}$ , $C=\sage {C1}$ , $\alpha =\sage {alpha1}$ and assuming that $\gamma$ is an acute angle.

However, since our right triangle tools don’t actually interact with this triangle, we will need to first add a line that can split this into two right triangles with a shared side. We accomplish this as follows:

Once we have this extra bit of information, we can see a couple of things. The angle $\beta$ has the same measurement as the sum of angles $\beta _1$ and $\beta _2$ and the line segment we called $B$ previously has the same length as the sum of $B_1$ and $B_2$ .

We can now throw away most of picture and just keep the triangle with sides $B_1$ , $C$ , and $D$ . Since we already know that $\alpha =\sage {alpha1}$ and $C=\sage {C1}$ we are left with finding all the sides of the following triangle:

We can now find:

$B1 = \answer {\sage {B1alpha}}$
$D = \answer {\sage {D1}}$
$\beta _1 = \answer {\sage {beta1alpha}}$

1.1 :

Now that we have those components we return to our larger picture and see that if we remove $B_1$ and $C$ we instead have the triangle:

Since we already know that $A=\sage {A1}$ and we just found that $D=\sage {D1}$ we can proceed to find:

$B_2 = \answer {\sage {B1gamma}}$
$\beta _2 = \answer {\sage {beta1gamma}}$
$\gamma = \answer {\sage {gamma1a}}$

1.1.1 : Now that we have those three pieces we can return to our original picture to find that we have almost everything. While we have found what $\gamma$ needs to be we have not yet found $\beta$ or $B$ . However, since $\beta _1+\beta _2=\beta$ and $B_1+B_2=B$ we can find the last two components of our original triangle as:

$\beta = \answer {\sage {beta1a}}$
$\gamma = \answer {\sage {gamma1a}}$

2 : Now we will find all of the missing angles and side lengths using the same $A$ , $C$ , and $\alpha$ as before, but assume that $\gamma$ is obtuse.

We previously added a line to split our triangle into two different right triangles, however if we were tr try the same thing we would have to split the angle $\gamma$ . We could proceed that way, but instead we are going to take our triangle and make it bigger. We do this by extending the side call $B$ until we can place our new side $D$ as follows:

As before, we have a right triangle in this picture that we know enough information about to find all of the pieces the triangle whose sides are $C$ , $D$ and $B'+B$ . We now throw away everything in the picture that isn’t a part of that triangle to get:

As before, since we know $C=\sage {C1}$ and $\alpha =\sage {alpha1}$ , we can use right triangle tools to find:

$B+B' = \answer {\sage {Bee}}$
$D = \answer {\sage {D2}}$
$\beta +\beta ' = \answer {\sage {BETA}}$

2.1 :

Since we now have $D=\sage {D2}$ and $A=\sage {A1}$ , we can now look back to our main picture and delete the line segment $B$ and the line segment $C$ to get:

We use this to find:

$B' = \answer {\sage {BPrime}}$
$\beta ' = \answer {\sage {betaprime}}$
$\gamma ' = \answer {\sage {gammaprime}}$

2.1.1 :

With all of those pieces found we return to our original picture to find that $\beta$ is just the difference of two of the angles we found and $B$ is just a difference of line segments. Similarly, we found $\gamma '$ , but what we need is $\gamma$ . This is less problematic than it might seem, though, since $\gamma +\gamma '$ is a straight line and so they are supplements. We can now state our remaining missing pieces as:

$B = \answer {\sage {B1b}}$
$\beta = \answer {\sage {beta1b}}$
$\gamma = \answer {\sage {gamma1b}}$