Question

a triangle has side lengths of 5,8, and 12. determine whether the triangle is right, acute or obtuse.

Answer 1

Acute

Answer 2

This triangle is obtuse.


Consider the triangle with side lengths \(a \text{, } b \text{, and } c\) and opposite angles \(A \text{, } B \text{, and } C\).

The Law of Cosines applies: \(c^2 = a^2 + b^2 - 2ab \cos C\)

Two cases (the third is already covered by Pythagorean Theorem):
\(\quad c^2 < a^2 + b^2\) if \(\neg 2ab \cos C < 0\)
\(\quad c^2 > a^2 + b^2\) if \(\neg 2ab \cos C > 0\)

Let's try to express this in terms of the angle only. We can divide off the \(\neg 2ab\) to simplify into terms of \(\cos C\) (flipping the sign, of course).

\(\quad c^2 < a^2 + b^2\) if \(\cos C > 0\)
\(\quad c^2 > a^2 + b^2\) if \(\cos C < 0\)

C is the angle opposite of the c-value. Consider the possible values of C, in terms of our triangle: \(0 < C < 180\).
We can then determine the values of C for which \(\cos C\) fits each of the two conditions:
\(\quad \cos C > 0\) when \(0 < C < 90\)
\(\quad \cos C < 0\) when \(90 < C < 180\)

Using this information, we can conclude:
. When C is an acute angle, then \(\cos C\) is greater than zero; therefore, \(c^2 < a^2 + b^2\) in this case.
. When C is an obtuse angle, then \(\cos C\) is less than zero; therefore, \(c^2 > a^2 + b^2\) in this case.

So, now let \(a = 5\), \(b = 8\), and \(c = 12\)
\( \quad 12^2 \ \_ \ 5^2 + 8^2 \\
\quad 144 \ \_ \ 25 + 64 \\
\quad 144 \ \_ \ 89 \\
\quad 144 > 89 \)
Therefore, this is \(c^2 > a^2 + b^2\); therefore, the triangle is obtuse.