Given the area of a triangle, 43560 square feet, and the interior
angles of the triangle (58 degrees 42 minutes 28 seconds, 59 degrees
45 minutes 15 seconds, and 61 degrees 32 minutes 17 seconds), what are
the lengths of the triangle?

Hi, kanesta:

Interesting problem!  Here's one approach.  We'll label the sides of
the triangle as a,b,c and the angles opposite them A,B,C respectively.

The triangle has sides of length b and c, with the included angle A
between them.  The area of the triangle is then:

area = (bc/2) * sin(A)

[It's well known, e.g see item 2 here:]

http://mathforum.org/library/drmath/view/59008.html

If you wrote the area for each pair of sides as above, and multiplied
the three equations together:

area^3 = (a^2)(b^2)(c^2) * [sin(A)sin(B)sin(C)/8]

abc = sqrt( 8(area^3)/[sin(A)sin(B)sin(C)] )

Now all the terms in the right hand side are known.  In particular:

sin(A)sin(B)sin(C) 
= 0.85452934585228245812168183086658 
* 0.86387213839760652692382353096153
* 0.8791338451317375067976646411563
= 0.64898020305672415111798738442483

so:

abc = 31919867.184934622063566119707661

and:

abc/[sin(A)sin(B)sin(C)] = 49184654.6852904

Recall from the law of sines that:

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

http://mathforum.org/library/drmath/view/51876.html

so that each of these ratios must be:

cuberoot(49184654.6852904) = 366.38966

Thus:

a = 366.38966 sin(A) = 313.0907 ft
b = 366.38966 sin(B) = 316.5138 ft
c = 366.38966 sin(C) = 322.10555 ft

As a quick check on our work, a perfect equilateral triangle of 1 foot
on each side would have area sqrt(3)/4 or roughly 0.433 sq. ft. 
Scaling that up by a factor of about 316^2 would give 0.433 * 316^2 =
43237.648.  This is close enough your actual area of 43560 sq. ft. to
assure us we have not made a "blunder" in the computation.

regards, mathtalk-ga

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Request for Answer Clarification by kanesta-ga on 15 Nov 2002 16:49 PST

Your answer to my question trig 101 was very good. I am a surveyor, and in
surveying we have an equation to solve this problem:

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

or 

a = SQ. RT. [(Area * 2 * sin A) / (sin B * sin C)]

By plugging in the area and angles, this equation makes the problem very
easy. 

Surveyors typically use this type of problem the write legal descriptions
for partitioning new parcels based on the clients requested area. 

Again, thank you for the answer.

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Clarification of Answer by mathtalk-ga on 15 Nov 2002 17:46 PST

Hi, kanesta-ga:

Thanks for taking time to rate my answer; it means a lot.  Also thanks
for explaining the context.  I was wondering how the area could be
known with precision (along with the angles) without knowing the
lengths in advance.  The application to creating subplots of
prescribed area makes perfect sense of this.

Let me add a few more words, to connect your solution with mine.  Your
equation:

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

can be rewritten, multiplying top and bottom by sin A:

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

from which one can deduce:

a / sin(A) = sqrt( 2 Area / (sin(A) sin(B) sin(C) )

This in essence is the approach I took, because the right hand side
above is used in common to express all three lengths a,b,c.

best wishes, mathtalk-ga