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### Finding a Missing Angle

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Date: 01/23/2002 at 21:11:07
From: Mel
Subject: Area of a triangle

I am home schooling and am stuck on an assignment about geometry, and
trigonometry. Here's one question:

I have a triangle with sides:

AB = 3.6 cm
BC = 6.5 cm
CA = 5.5 cm
and angle CAB is 90 degrees.

The question is: Using trigonometry, calculate the measure of angles
ABC and ACB.

How can I find the missing angle? If I had two angles I could subtract
them from 180 degrees, but I only have one.

And I don't know how to find the area of a triangle. In my assignment
a triangle measures:

CA = 10.8cm
AB = 7.2cm
BC = 13cm
and it says find the area.  But I don't know how.

Can you help me?

Mel
```

```
Date: 01/24/2002 at 20:11:06
From: Doctor Jeremiah
Subject: Re: Area of a triangle

Hi Mel,

Trigonometry is the way you find angles when you know only one angle.
There are only three important things you need to know to do
trigonometry. First, the triangle has to have a 90-degree angle.
Second, you need to know the definitions for sine, cosine, and
tangent. And third, in order to make those definitions you need to
define a couple of other things: the hypotenuse, the adjacent side,
and the opposite side.

The triangle MUST have a 90-degree angle. And what good luck, yours
does!
C
+
/|
/ |
/  |
/   |
/    |
6.5   5.5
/      |
/       |
/        |
+---3.6---+
B           A

The hypotenuse is the side that does not touch the 90-degree angle.
It is also the longest side. In your triangle it is side BC and has a
length of 6.5

The adjacent side and the opposite side move around depending on what
angle you want to calculate. Say you want to find the value of angle
CBA.

The adjacent side is the side that touches angle CBA (and isn't the
hypotenuse). In this case it is side AB because side AB touches angle
CBA and side AC does not.

The opposite side is the side that does not touch angle CBA (and also
isn't the hypotenuse). In this case it is side AC because side AC does
not touch angle CBA and side AB does.

And if you are trying to find angle BCA then BC is the hypotenuse,
AB is the opposite side, and AC is the adjacent side.

So what are these definitions good for? Well now we can define the
sine, cosine,  and tangent:

sine: sin(angle) = opposite/hypotenuse

Lets say you want to find the size of angle CBA. The hypotenuse is BC,
the adjacent side is AB, and the opposite side is AC.

We choose an equation from among sine, cosine, and tangent depending
on what information we know. If we know only two sides, we will pick
the equation that uses those two sides. But we know all three sides so
it doesn't matter.

So we will pick one:

sin(angle) = opposite/hypotenuse
sin(CBA) = AC/BC
sin(CBA) = 5.5/6.5
sin(CBA) = 0.846

Now how do we change this into something that equals angle CBA? Right
now it's the sine of CBA and that is not useful.

Well, you know how the square and square root cancel each other out:
sqrt(x squared) = x. It turns out that there are also functions that
undo sine, cosine, and tangent. They are called the arcsine,
arccosine, and arctangent:

arcsine: arcsin( sin(angle) ) = angle

arccosine: arccos( cos(angle) ) = angle

arctangent: arctan( tan(angle) ) = angle

So when we have an equation that looks like:

sin(CBA) = 0.846

We can do this:

sin(CBA) = 0.846
arcsin( sin(CBA) ) = arcsin( 0.846 )
CBA = arcsin( 0.846 )

Now, to do the arcsin you will need a calculator or a table or
arcsines or a slide rule.

On a calculator the arcsine is sometimes called the inverse sine, or
sometimes has the symbol of sin with a -1 exponent, and sometimes it's
unlabeled but is on the secondary function ogf the sine button.

When you don't know for sure that the triangle has a 90-degree angle,
you have to be sneaky. If you know for sure that it has a 90-degree
angle, then it's just half a rectangle (cut diagonally). But without
a 90-degree angle we need to use Heron's formula.

Heron also proves his famous formula:

If A is the area of a triangle with sides a, b, and c, and
s = (a+b+c)/2, then A = sqrt[s(s-a)(s-b)(s-c)].

Let's change your triangle to use sides named a, b and c:

a = 7.2cm
b = 10.8cm
c = 13cm

Now:

s = (a+b+c)/2
s = (7.2+10.8+13)/2
s = 15.5

And:

A = sqrt[ s(s-a)(s-b)s-c) ]
A = sqrt[ 15.5 (15.5 - 7.2) (15.5 - 10.8) (15.5 - 13) ]
A = sqrt[ 15.5 (8.3) (4.7) (2.5) ]
A = 38.87978

Let me know if you still have questions.

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

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