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The given problem involves using the cosine rule to find the angle \( A \) in a triangle with sides of lengths 4, 5, and 6 units.

The cosine rule (law of cosines) states:
\[
\cos(A) = \frac{b^2 + c^2 - a^2}{2bc}
\]
where:
- \( A \) is the angle we want to find,
- \( a \) is the side opposite to angle \( A \) (which is 5 units),
- \( b \) and \( c \) are the other two sides (4 and 6 units).

### Step-by-Step Solution:

1. Using the cosine rule:
\[
\cos(A) = \frac{4^2 + 6^2 - 5^2}{2 \times 4 \times 6}
\]
\[
\cos(A) = \frac{16 + 36 - 25}{48}
\]
\[
\cos(A) = \frac{27}{48} = 0.5625
\]

2. Now, to find angle \( A \), we take the inverse cosine (cosine inverse) of 0.5625:
\[
A = \cos^{-1}(0.5625)
\]

Using a calculator:
\[
A \approx 55.771^\circ
\]

Thus, the angle \( A \) is approximately \( 55.771^\circ \) when rounded to three decimal places.

Would you like further details or have any questions? Here are 5 related questions to deepen understanding:

1. How can the cosine rule be applied to find other angles in a triangle?
2. What is the sine rule, and how is it different from the cosine rule?
3. How would you solve this problem if all angles were given, but only one side length was missing?
4. Can the cosine rule be used for obtuse triangles, and how would that affect the calculation?
5. What is the significance of rounding to three decimal places in trigonometric calculations?

**Tip:** When using the cosine rule, ensure the side lengths correspond correctly to the angle you are solving for to avoid errors in the calculation.

Use the cosine rule to find the value of angle A in a triangle with sides of lengths 4, 5, and 6. Give your answer correct to three decimal places.

Learn how to use the cosine rule to find the value of angle A in a triangle with sides 4, 5, and 6. This trigonometry problem demonstrates the law of cosines step-by-step, providing a clear solution for finding the angle, rounded to three decimal places.

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