Question

Use the Law of
Cosines, a'=62 + c* - 2bc• cos A, to find the value of angle A in a right
triangle, where a = 9, 6 = 40, and
c =41. (1 point)
• 13.1 degrees
• 12.7 degrees
• 17.3 degrees
• 14.5 degrees

Answer 1

To find the angle \( A \) in a right triangle using the Law of Cosines, we start with the formula given:

\[
a' = b^2 + c^2 - 2bc \cdot \cos A
\]

We need to rearrange it to solve for \( \cos A \):

\[
\cos A = \frac{b^2 + c^2 - a'}{2bc}
\]

Here, we have the following values:
- \( a' = 9 \)
- \( b = 40 \)
- \( c = 41 \)

Now substituting those values into the formula:

1. Calculate \( b^2 \) and \( c^2 \):
\[
b^2 = 40^2 = 1600
\]
\[
c^2 = 41^2 = 1681
\]

2. Now substitute these into the rearranged Law of Cosines formula:
\[
\cos A = \frac{1600 + 1681 - 9}{2 \cdot 40 \cdot 41}
\]
\[
\cos A = \frac{1600 + 1681 - 9}{3280}
\]
\[
\cos A = \frac{3272}{3280}
\]

3. Simplify \( \cos A \):
\[
\cos A = 0.996
\]

4. Now calculate \( A \):
\[
A = \cos^{-1}(0.996) \approx 8.1^\circ
\]

It appears that my calculations are unintentionally leading toward an angle that doesn’t match the options provided. Let's recheck:

- Inserting values into \( A = \cos^{-1}( \frac{b^2 + c^2 - a^2}{2bc} ) \):

5. Calculation:
\[
\cos A = \frac{1600 + 1681 - 81}{3280}
\]
\[
\cos A = \frac{3200}{3280}
\]
\[
\cos A = 0.9756
\]

Thus,
\[
A \approx \cos^{-1}(0.9756) \approx 12.346^\circ
\]

It seems I had some prior errors, correcting now seems like \( \cos A \) should be placed correctly, finding approximate:

Now let’s select the closest angle option you provided from previously calculated results:

- **Closest matching given values would be \( 12.7^\circ \)**.

Therefore, the answer is **12.7 degrees**.