Expert: Paul Klarreich - 2/26/2006

Question
how would I find the area of an triangle with a=11.8, b=12.6, and c=14.8

Answer
Hi, Dipti,

Subject:  Cal/ Trig
Question:  how would I find the area of an triangle with a=11.8, b=12.6, and c=14.8

>> In the future, if you have two questions to ask, just put them together.  There is a limit on the number of questions they will send through, so you use up that limit, and someone else won't get to ask.

There is an easy formula for the area of a triangle.  If you have two sides and the angle between them (you heard that before), if the two sides are called a, b, and the angle is C, then

Area = 1/2 a b sin C.

In this case, you don't have any angle, so you have to find one.  Use the

L-o-C again, to find angle C:

c^2 = a^2 + b^2 - 2 a b cos C

Put:

(14.8)^2 = (11.8)^2 + (12.6)^2 - 2(11.8)(12.6) cos C

Now solve that for angle C.  You will have:

[VIEW IN A FIXED-SIZE FONT, LIKE COURIER.]

       (14.8)^2 - (11.8)^2 - (12.6)^2       
cos C = -------------------------------
            - 2(11.8)(12.6)

or
         (11.8)^2 + (12.6)^2 - (14.8)^2
cos C =  --------------------------------
               2(11.8)(12.6)

Then use the arc cos  key to get the value of C, in degrees.  [On the Windows calculator you must use the INV box plus the cosine key.]

Once you have that, use it in the area formula above.

NOTE: If you don't like this, there is something called Hero's formula.  You can easily look it up.