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### Area of a Triangle

```
Date: 6/2/96 at 19:53:54
From: Anonymous
Subject: Area of a triangle

Hello,

How can I calculate the area of a triangle if I know that one of its
sides measures l and that its two adjacent angles measure a and b?

I have tried many things, and I feel that I should use trigonometry,
though I am not sure how. Maybe I should think about a parallelogram
and the congruent angles formed by its diagonals?
```

```
Date: 6/3/96 at 9:33:39
From: Doctor Pete
Subject: Area of a triangle

/|\
/ |  \
/  |    \
/   |h     \
/    |        \
a /_____|__________\ b
x     l-x

so the sum of the base is x+l-x = l, your given length.

Then tan(a) = h/x, and tan(b) = h/(l-x),
so h = x tan(a) = (l-x) tan(b) ==> x(tan(a)+tan(b)) = l tan(b)
==> x = l tan(b)/(tan(a)+tan(b)).

Hence h = x tan(a) = l*tan(a)tan(b)/(tan(a)+tan(b)). Then the area of
the triangle is:

h*l   l^2*tan(a)tan(b)
--- = ---------------- .
2    2(tan(a)+tan(b))

Alternatively, this can be rewritten to include the third angle c,
which is Pi-(a+b) radians or 180-(a+b) degrees:

l^2 /       1                \
Area = --- | ------------- + cot(c) | ,
2  \ tan(a)+tan(b)          /

which might be easier to use.

-Doctor Pete,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

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