1. Law of Sines:
•Formula:
• Why do we need it?
- The law of sines is very useful
for solving triangles, especially non-right triangles. Right triangles are easy
to solve for because one can just use Pythagorean Theorem or the normal trig
functions to find the missing lengths or angles, but those theorems and
functions can’t be used to solve non-right triangles. The law of sines works
for any triangle.
• How is it derived from what we
already know?
1. Let’s say we have this non-right triangle
like the one below:
2.
Let’s make a perpendicular line to side AC from
angle B. Label it as h.
3. We now have made 2 right triangles, so we can
use the normal trig functions (sin, cos, tan, etc.) to determine the
relationships between the sides and angles. For example, sinA= h/c, and sinC=
h/a.
4. Now, when we look at the two sin ratios above,
we notice that they both have h as their common variable. So, we’ll simply both
equations by multiplying the denominators to both sides of their respective
equations to get them both equal h and therefore each other to make
csinA=asinC.
5. Divide both sides by the coefficients of a and c
and you are left with:
6. You’ll probably be wondering: “Well, what about
angle B?” Because we made a perpendicular line from angle B, cutting it into 2
pieces, we can’t use the trig functions for angle B. However, if we made a
perpendicular line (h) from another angle, say A.
7. We have again 2 right triangles so
we can use the trig functions for angle B now, which would be sinB=h/c and
angle C would be sinC=h/b.
8. Simplify like we did before (multiply denominators
to make equations equal to h and therefore each other). And you’ll end up with:
9. And there is your law of sines:
4. Area of an Oblique
Triangle:
• Definition: a triangle with all
of its sides of different lengths
• Formula: Area= ½ bcsinA or Area= ½ acsinB or Area= ½
absinC (depending on which angle you
are given
• How is
the area of an oblique triangle derived?
1. Below is an oblique triangle of
which it is difficult to calculate the height of normally. So we add h as a
line that would make the triangle into a right triangle, where we can use the
normal trig functions.
2. Since we can use the normal
trig functions, we know that sinC=h/a, sinA=h/c, and sinB=h/(a or c) if h
dropped down from a different angle (like with deriving the law of sines
above).
3. Then, we simplify these
equations again by multiplying them by their denominators to make them equal to
h.
4. Now instead of equaling them to
each other (since that won’t help us calculate the area), we plug each of them
into the formula for the area of a triangle (A= ½ bh).
5. So h would be replaced by asinC,
or csinA. That would be how the formula becomes A= ½ b(csinA) or A= ½ a(bsinC).
If you do the same thing for the other triangle to find B, then it ends up as
A= acsinB.
• How does it relate to the area
formula that you are familiar with?
- As we saw how the area of an oblique
triangle is derived, we see that we have used the normal formula for the area
of a triangle. We plugged in the values into h and made new formula. However,
one should keep in mind that the area of an oblique triangle will only work if
you have two side lengths and their included angle. The angle has to be between
the two sides so they can’t have the same letters basically.
References:
