finding area of triangle

cheetahbelly13

New member
Joined
Apr 22, 2006
Messages
6
i need help finding the area of a triangle with sides of 16,20 with an angle of 30 degrees between them. ive drawn a picture but i dont know how to put it on here.
 
What tools do you get to use?

The Law of Cosines will give you the other side.
Sines and Cosines by themselves will give altitudes.

Lots of ways to go on this one.
 
If a & b are the lengths to two sides of a triangle and Θ is the angle between them then the area of the triangle is
\(\displaystyle \L
A = \frac{{a \cdot b \cdot \sin (\theta )}}{2}\).
 
Hello, cheetahbelly13!

Find the area of a triangle with sides of 16,20 and an angle of 30 degrees between them.
pka gave you the formula for this problem.
\(\displaystyle \;\;\)With a bit of imagination, you could have invented it yourself.
Code:
            *
           /: \
          / :   \
        a/  :     \
        /   :h      \
       /θ   :         \
      * - - - - - - - - *
              b
We are given two sides of a triangle, \(\displaystyle a\) and \(\displaystyle b\),
\(\displaystyle \;\;\) and the included angle, \(\displaystyle \theta.\)

The area of a triangle is: \(\displaystyle \,A\;=\;\frac{1}{2}bh\) [1]

In the diagram, we see that: \(\displaystyle \,sin\theta\,=\,\frac{h}{a}\;\;\Rightarrow\;\;h\,=\,a\cdot\sin\theta\)

Substitute into [1]: \(\displaystyle \,A\;=\;\frac{1}{2}b(a\cdot\sin\theta)\)

Therefore: \(\displaystyle \L\:A\;=\;\frac{1}{2}ab\sin\theta\)
 
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