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Area of Quadrilateral, Given Angles and Two Opposite Sides

Date: 05/13/2002 at 13:23:20
From: Simon Murtha-Smith
Subject: Find the area of quadrilaterals with specific side and 
angle configurations

My question is this: If I have a quadrilateral with all the 
angles known and two opposite sides known, can I find the area?

I have tried splitting the shape in two to form two triangles, 
each with one side and one angle known, and then using the law of 
sines and the law of cosines; but I end up stuck with an angle 
inside of a sin or cos and cannot get it out. 

This is my work so far. A quadrilateral with known, opposite 
sides, a and c, and angles A, B, C, and D might look something 
like this

|                  |
|                  |
|                  |
c                  a
|                  |
|                  |
|                  |

The angles are not necessarily 90 degree angles.

I then draw the diagonal

| \                |
|   \              |
|      \           |
c         e        a
|           \      |
|              \   |
|                \ |

I now solve for the angles formed by the two triangles.

| \180-(C-x)       |
|    \             |
|180-D-x\          |
c         e        a
|           \      |
|              \C-x|
|              x \ |

Now using the law of sines I make the following equations:

  e/sin B = a/sin(180-C+x)

  e = -a*sin B / sin(x-C)


  e/sin D = c/sin(180-D-x)


  e = c*sin D / sin(x+D)

from these I get

  sin(x+D)/sin(x-C) = ( c*sin D )/( a*sin B )

I can then expand the sines on the left side, but it is useless. 
I end up with

 ((sin x)(cos D) + (cos x)(sin D))
 ((sin x)(cos C) - (cos x)(sin C))

I can't go any further. There doesn't seem to be a way to pull 
the x out in order to solve for all the angles and then solve for 
the whole quadrilateral. 

I really need help on this one. I am so close but I can't it 
solved. Thanks for your help.

Simon Murtha-Smith

Date: 05/13/2002 at 16:04:08
From: Doctor Rick
Subject: Re: Find the area of quadrilaterals with specific side 
and angle configurations

Hi, Simon.

You've done good work so far. I don't see another approach, and 
this one should work in principle, so let's try to take it 

Cross-multiplying (that is, multiplying through by both 
denominators), we have

  a*sin(B)*sin(x+D) = c*sin(D)*sin(x-C)

which can be expressed using the angle-sum formulas as

  a*sin(B)(sin(x)*cos(D)+cos(x)*sin(D)) =

We want to solve this equation for x. The best way to do so is to 
rewrite the equation so it involves only one trig function of x. 
Solving this equation for the trig function of x, we'll be one 
step from a solution for x.

I'll do a bit of rearrangement to reduce the number of times x 
appears; this should give you a good hint.

   sin(x)(a*sin(B)*cos(D)-c*sin(D)*cos(C)) =

Let us know what you come up with; this could be good material 
for our archives. Check my work, I'm doing it quickly!

- Doctor Rick, The Math Forum 

Date: 05/13/2002 at 19:52:21
From: Simon Murtha-Smith
Subject: Find the area of quadrilaterals with specific side and 
angle configurations

Thank you very much Dr. Rick.

That was very useful. I managed to get the answer in terms of x. 
This is what I continued to do from where you left off.

You wrote:

   sin(x)(a*sin(B)cos(D) - c*sin(D)cos(C)) =
  -cos(x)(a*sin(B)sin(D) - c*sin(D)sin(C))

I then divided by cos(x) to get

   tan(x)(a*sin(B)cos(D) - c*sin(D)cos(C)) =
  -(a*sin(B)sin(D) - c*sin(D)sin(C))

I then brought everything but the tan(x) over to get

           -(a*sin(B)sin(D) - c*sin(D)sin(C)) 
  tan(x) = ----------------------------------
           (a*sin(B)cos(D) - c*sin(D)cos(C))

Then I just took the arctan to get 

              -(a*sin(B)sin(D) - c*sin(D)sin(C)) 
  x = arctan( ---------------------------------- )
               (a*sin(B)cos(D) - c*sin(D)cos(C))

Thanks for all your help. I am working on a project on finding 
the area of irregular polygons using trigonometry and this was 
one particular case of a polygon that I knew I could solve for 
the area but wasn't sure how to finish the problem.
Associated Topics:
High School Triangles and Other Polygons
High School Trigonometry

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