Question

Use the stated variant of the Law of Cosines, a^2=b^2+c^2-2bc · cos A , to find the value of angle A in a right triangle, where a=9, b=40, c=41.

Answer 1

In a right triangle, we can use the Law of Cosines to find the angle \( A \) opposite side \( a \). We are given:

- \( a = 9 \)
- \( b = 40 \)
- \( c = 41 \)

The Law of Cosines states:

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

We replace \( a \), \( b \), and \( c \) into the formula:

\[
9^2 = 40^2 + 41^2 - 2 \cdot 40 \cdot 41 \cdot \cos A
\]

Calculating the squares:

- \( 9^2 = 81 \)
- \( 40^2 = 1600 \)
- \( 41^2 = 1681 \)

Substituting these values:

\[
81 = 1600 + 1681 - 2 \cdot 40 \cdot 41 \cdot \cos A
\]

Calculating \( 1600 + 1681 \):

\[
81 = 3281 - 2 \cdot 40 \cdot 41 \cdot \cos A
\]

Calculating \( 2 \cdot 40 \cdot 41 \):

\[
2 \cdot 40 \cdot 41 = 3280
\]

Substituting this result back into the equation:

\[
81 = 3281 - 3280 \cdot \cos A
\]

Now, isolate \( \cos A \):

\[
3280 \cdot \cos A = 3281 - 81
\]
\[
3280 \cdot \cos A = 3200
\]
\[
\cos A = \frac{3200}{3280}
\]

Simplifying the fraction:

\[
\cos A = \frac{320}{328} = \frac{40}{41}
\]

Now, we find angle \( A \) by taking the inverse cosine:

\[
A = \cos^{-1}\left(\frac{40}{41}\right)
\]

Using a calculator:

\[
A \approx 18.92^\circ
\]

Thus, the value of angle \( A \) is approximately \( 18.92^\circ \).