Question

How do you find the area of the triangle ABC with a = 7.2, c = 4.1, and B=28 degrees?

Answer 1

The area of the triangle is approximately `6.93`.

Explanation:

For this problem, we will use the following:
The law of cosines:
`c^2 = a^2 + b^2 -2abcos(C)`

Heron's formula:
`Area = sqrt(p(p-a)(p-b)(p-c))` where `p = (a+b+c)/2`

Applying the law of cosines gives us

`b^2 = a^2 + c^2 - 2accos(B)`

`=> b^2 = 7.2^2 + 4.1^2 - 2(7.2)(4.1)cos(28^@)`

`=> b = sqrt(51.84 + 16.81 - 2(7.2)(4.1)cos(28^@)) ~~ 4.065`
(Note that for greater accuracy, we can perform the calculation of the square root at the very end, however for convenience we will use this approximation here)

Now, let `p = (a+b+c)/2 ~~ 7.6825`

By Heron's formula, then, we get the area `A` of the triangle as

`A = sqrt(p(p-7.2)(p-4.065)(p-4.1))`

Plugging this in gives us the final result

`A ~~ 6.93`