Question

Triangle ABC is shown. What is cos (angle BAC)?
Side AB: 5
Side BC: 3
side AC: 4

Answer 1

To find \(\cos(\angle BAC)\) in triangle \(ABC\) with sides \(AB = 5\), \(BC = 3\), and \(AC = 4\), we can use the Law of Cosines. The Law of Cosines states that for any triangle:

\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]

In triangle \(ABC\), we can designate:

- \(a = BC = 3\)
- \(b = AC = 4\)
- \(c = AB = 5\)
- \(\angle C = \angle BAC\)

Substituting the values into the Law of Cosines formula, we get:

\[
5^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cdot \cos(\angle BAC)
\]

Calculating the squares:

\[
25 = 9 + 16 - 24 \cdot \cos(\angle BAC)
\]

Simplifying the right side:

\[
25 = 25 - 24 \cdot \cos(\angle BAC)
\]

Now, we isolate \(\cos(\angle BAC)\):

\[
25 - 25 = -24 \cdot \cos(\angle BAC)
\]

This simplifies to:

\[
0 = -24 \cdot \cos(\angle BAC)
\]

From this, we have:

\[
\cos(\angle BAC) = 0
\]

This means that \(\angle BAC\) is \(90^\circ\). In conclusion, the value of \(\cos(\angle BAC)\) is:

\[
\boxed{0}
\]