The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Area of a Triangle

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


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

Say you had

        / |  \
       /  |    \
      /   |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!   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.