Download A couple harder examples of triangle solving are found here.

Additional Exercises for Sections 7.1 and 7.2
In this document, we will show off two problem types that use the material of Section 7.1 and Section 7.2.
In the first, we’ll show how you can solve a triangle when you know all its sides (the SSS scenario). In the
second, we’ll deal with an unusual setup with an ASA scenario (side between two angles), where we have to get
the height of a triangle.
Solving an SSS triangle
Ex: Solve the triangle with the given information: a = 10, b = 15, and c = 22.
We know all the sides, so this is an SSS scenario. We need to find the angles. When you know a lot of sides,
the Law of Cosines (LOC) is appropriate.
1. Since we know all the sides, we can find any angle we want first via LOC. I recommend finding the largest
angle first, opposite the largest side, so we’ll go with γ. We’ll use the version of LOC which uses the γ, c
pair:
c2 = a2 + b2 − 2ab cos(γ)
We want to know γ, so let’s solve for cos(γ) on its own:
c2 − a2 − b2 = −2ab cos(γ)
⇒
cos(γ) =
c2 − a2 − b2
−2ab
Lastly, we use the inverse cosine and plug in our known values:
2
22 − 102 − 152
γ = arccos
≈ 122.01◦
−2(10)(15)
(Use degrees mode on your calculator to get this.)
2. Next, let’s get β (though you could just as easily get α instead). You can either do a similar approach to
Step 1, or you can use the fact that Step 1 now gave you a completed γ, c pair for use in the Law of Sines
(LOS). I think the second approach is easier; we’ll use LOS with the opposing pairs β, b and γ, c:
sin(γ)
sin(β)
=
b
c
If you multiply b and then use the inverse sine, you get
b sin(γ)
15 sin(122.01◦ )
β = arcsin
≈ arcsin
≈ 35.32◦
c
22
3. Lastly, the easiest way to get α is to take the other two angles and subtract from 180◦ :
α = 180◦ − β − γ ≈ 180◦ − 35.32◦ − 122.01◦ ≈ 22.67◦
The approximate final answer is α ≈ 22.67◦ , β ≈ 35.32◦ , γ ≈ 122.01◦ .
Note how α is the smallest angle, opposite the smallest side a, and similarly γ is the largest angle opposite
the largest side c.
Height of an ASA Triangle with a Variable Base
Ex: A triangle has base c and base angles 36◦ and
79◦ . Determine the height as a function of c.
Warning: If you draw the height to split this into
right triangles, the two triangles are not congruent!
In other words, the base does not split into c/2 and
c/2.
Hint: Find some other side of the triangle first in
terms of c.
Brainstorming: There’s quite a lot to do in this problem. However, the first step should be fairly clear;
try drawing a height in the triangle and labeling it h. That will split up our triangle into two right triangles.
However, as warned, the two triangles are not congruent; we cannot assume the two parts are equal.
Let’s say the lower-left corner is A, the lower-right
is B, and the top is C. When we draw our height, it
strikes the base at some point we’ll call D. This gives
us two right triangles ∆ADC and ∆BDC, as well as
the biggest triangle ∆ABC that we started with.
At the moment, we don’t know any sides at all of our right triangles, so we can’t solve them. However, we
can get one of their hypotenuses (either b = AC or a = BC) by using our techniques from 7.1 and 7.2 on the
big triangle ∆ABC! (It will be an answer in terms of c, but that’s acceptable for this problem.) Once we’ve
done that, we’ll have a right triangle with a known hypotenuse and acute angle, so we can finish the problem
with SOHCAHTOA.
Answer: As described in the brainstorming, let’s label a height on this triangle and get points A, B, C, and
D marked. Let h be the height of the triangle ∆ABC. We notice we have an ASA scenario with the angles of
α = 36◦ , β = 79◦ , and the side c.
1. First, we find the angle γ = ∠ACB by γ = 180◦ − α − β = 180◦ − 36◦ − 79◦ , so γ = 65◦ .
2. Now we have the opposing pair γ, c in ∆ABC, so let’s figure out the side a = BC from the pair α, a by
using LOS. We find
c
c sin(α)
c sin(36◦ )
a
=
⇒ a=
=
sin(α)
sin(γ)
sin(γ)
sin(65◦ )
3. By finding a, we found the hypotenuse of the right triangle ∆BDC. We want the height h opposite the
angle β = 79◦ , so we use
opp
h
sin(79◦ ) =
=
hyp
a
Multiplying a to both sides and plugging in the value from Step 2, we finally get the answer:1
h=
c sin(36◦ ) sin(79◦ )
sin(65◦ )
Remark: Notice that this answer needed to use every angle from the triangle! It also needed to use that
side of c; that’s because knowing all angles (an AAA scenario) is not enough to give you any side lengths of a
triangle.
1 In WebAssign, you may see the answer key write the denominator as sin(115◦ ) instead of sin(65◦ ). It turns out these two
values are the same; there’s a property about supplementary angles and sines that you don’t have to worry about. I guess that
WebAssign prefers to use α + β rather than to subtract that sum from 180◦ to get γ properly...