Area of a Triangle

Date: 10/18/2003 at 04:29:10
From: Kevin
Subject: The Area of a Triangle

Why can the area of a triangle be calculated using the formula

   K = a^2*sin(B)*sin(C)/[2*sin(B+C)] ?

Date: 10/18/2003 at 11:17:05
From: Doctor Luis
Subject: Re: The Area of a Triangle

Hi Kevin,

Please look at the following triangle.

  

The area K of that triangle is

  K = (1/2)h*b

So, to find the area, we need to find h and b.

From trigonometry, we know that h = a sin(C).  We can get b from the
law of sines:

  sin(B)/b = sin(A)/a

         b = a*sin(B)/sin(A)

Therefore,

  K = (1/2)(a*sin(C))*(a*sin(B)/sin(A))

    = a^2 sin(B)sin(C)/(2*sin(A))

We're almost there.  If you remember from geometry, the internal 
angles of a triangle sum to 180 degrees, or pi radians.

  A + B + C = pi

          A = pi - (B + C)

Also,

  sin(A) = sin(pi - (B + C))

         = sin(pi)cos(B+C) - cos(pi)sin(B+C)

         = 0*cos(B+C) - (-1)*sin(B+C)

         = sin(B + C)

And that pretty much proves it:

  K = a^2*sin(B)*sin(C)/[2*sin(B + C)]

I hope this helped!  Let us know if you have any more questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/