Three sides of a triangle:Area from Heron's formula, angles from the Law of Cosines

The Law of Cosine has three different statements, given side lengths a, b and c and angles opposite alpha, beta and gamma, respectively.

a² = b² + c² - 2bccosalpha
b² = a² + c² - 2accosbeta
c² = a² + b² - 2abcosgamma

On earlier homework, we figured out the third side from two lengths and the measure of the angle between them.  With a little algebraic manipulation, we can figure out the cosines of the three angles is we are given the three side lengths, and using the cosine inverse function, written here as arccos(number), we can find the three angles, or at least approximations, since the angles will often be given as irrational numbers.

cosalpha = [b² + c² - a²]/[2bc]
cosbeta = [a² + c² - b²]/[2ac]
cosgamma = [a² + b² - c²]/[2ab]

Examples

If we know the perimeter of a triangle is 7 and the sides are all whole numbers, there are only two possible answers, <3, 3, 1> and <3, 2, 2>. Let's answer all the questions from the quiz base on these numbers

c = 3, b = 3, a = 1
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perimeter = 7, so semi-perimeter = 7/2

The triangle is isosceles and acute.

Area = sqrt(7/2[7/2 - 3][7/2 - 3][7/2 - 1]) = sqrt(7/2 × 1/2 × 1/2 × 5/2) = sqrt(35/16) = sqrt(35)/4
approximation: 1.4790

Because c = b, gamma = beta, so we only need to do one calculation to find both.
cosgamma = [1 + 9 - 9]/[2 × 1 × 3] = 1/6
arccos(1/6) ~= 80.4059°
So gamma = beta ~= 80.4059°

If we calculate alpha, we get cosalpha = [9 + 9 - 1]/[2 × 3 × 3] = 17/18
arccos[17/18] = 19.1881°
So alpha ~= 19.1881°

You get the same answer if you subtract 2gamma from 180°.

--
c = 3, b = 2, a =2
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perimeter = 7, so semi-perimeter = 7/2

The triangle is isosceles and obtuse.

Area = sqrt(7/2[7/2 - 3][7/2 - 2][7/2 - 2]) = sqrt(7/2 × 1/2 × 3/2 × 3/2) = sqrt(63/16) = 3sqrt(7)/4
approximation: 1.9843

Because a = b, alpha = beta, so we only need to do one calculation to find both.
cosalpha = [9 + 4 - 4]/[2 × 3 × 2] = 9/12 = 3/4
arccos(1/6) ~= 80.4059°
So alpha = beta ~= 41.4096°

If we calculate gamma, we get cosgamma = [4 + 4 - 9]/[2 × 2 × 2] = -1/8
arccos[-1/8] = 19.1881°
So gamma ~= 97.1808°

You get the same answer if you subtract 2alpha from 180°.

Practice problems posted on Wednesday.

Find the classifications, exact and approximate area and approximate all three angles of the following triangles with perimeter = 11.

a) side lengths c = 5, b = 5, a = 1
b) side lengths c = 5, b = 4, a = 2
c) side lengths c = 5, b = 3, a = 3

Answers in the comments.