Question

What is the area of a triangle with sides of length 4, 6, and 9?

Answer 1

Let a = 4, b = 6, and c = 9. `A=cos^-1((6^2+9^2-4^2)/(2*6*9))=20.74...`
`Area=1/2 b c sin A=1/2(6)(9)sin 20.74... = 9.56 unit^2`

Explanation:

First we let a=4, b=6,c=9. Then we need to find one of the angles. I found A but you could find B or C. Note that whatever angle you are solving for its side will only appear after the minus sign. Once you find the angle then plug it in to the area formula and whatever angle you are using you may not use its side in the calculation.