Question
Triangle ABC is shown. What is cos (angle BAC)?
Side AB: 5
Side BC: 3
side AC: 4
A. 5/4
B. 3/4
C. 3/5
D. 4/5
Side AB: 5
Side BC: 3
side AC: 4
A. 5/4
B. 3/4
C. 3/5
D. 4/5
Answers
GPT-4o mini
To find \(\cos(\angle BAC)\) in triangle ABC, we can use the Law of Cosines. The Law of Cosines states that for any triangle, the relationship between the lengths of the sides and the cosine of one of the angles is given by:
\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]
where \(c\) is the length of the side opposite angle \(C\), and \(a\) and \(b\) are the lengths of the other two sides.
In triangle ABC, let:
- \(AB = c = 5\)
- \(AC = b = 4\)
- \(BC = a = 3\)
We want to find \(\cos(\angle BAC)\), which corresponds to \(\angle A\) in our setup. We can apply the Law of Cosines as follows:
\[
AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(\angle A)
\]
Substituting in the side lengths:
\[
5^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3 \cdot \cos(\angle A)
\]
Calculating the squares:
\[
25 = 16 + 9 - 24 \cdot \cos(\angle A)
\]
Combining the terms on the right:
\[
25 = 25 - 24 \cdot \cos(\angle A)
\]
Subtracting 25 from both sides:
\[
0 = -24 \cdot \cos(\angle A)
\]
Since \(-24 \cdot \cos(\angle A) = 0\), it follows that:
\[
\cos(\angle A) = 0
\]
This means that \(\angle A\) is \(90^\circ\), and we realize that in this case the cosine isn't calculated correctly. After reviewing my calculations, everything is in fact accurate. Here, side measurements indicate that \(AB\) wasn't actually used adequately.
This leads to:
\[
24 \cos(\angle A) = 0 → \text{a mistake in cosine components}
\]
The remaining method will use ratios and by the calculation yields \( cos(a) = \frac{adjacent}{hypotenuse} = \frac{4}{5}\) Finally after recapping terms and arrangements I confirm:
Hence, the answer is:
\(\boxed{\frac{4}{5}}\)
\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]
where \(c\) is the length of the side opposite angle \(C\), and \(a\) and \(b\) are the lengths of the other two sides.
In triangle ABC, let:
- \(AB = c = 5\)
- \(AC = b = 4\)
- \(BC = a = 3\)
We want to find \(\cos(\angle BAC)\), which corresponds to \(\angle A\) in our setup. We can apply the Law of Cosines as follows:
\[
AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(\angle A)
\]
Substituting in the side lengths:
\[
5^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3 \cdot \cos(\angle A)
\]
Calculating the squares:
\[
25 = 16 + 9 - 24 \cdot \cos(\angle A)
\]
Combining the terms on the right:
\[
25 = 25 - 24 \cdot \cos(\angle A)
\]
Subtracting 25 from both sides:
\[
0 = -24 \cdot \cos(\angle A)
\]
Since \(-24 \cdot \cos(\angle A) = 0\), it follows that:
\[
\cos(\angle A) = 0
\]
This means that \(\angle A\) is \(90^\circ\), and we realize that in this case the cosine isn't calculated correctly. After reviewing my calculations, everything is in fact accurate. Here, side measurements indicate that \(AB\) wasn't actually used adequately.
This leads to:
\[
24 \cos(\angle A) = 0 → \text{a mistake in cosine components}
\]
The remaining method will use ratios and by the calculation yields \( cos(a) = \frac{adjacent}{hypotenuse} = \frac{4}{5}\) Finally after recapping terms and arrangements I confirm:
Hence, the answer is:
\(\boxed{\frac{4}{5}}\)