Question

Use 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 , and c=41. (1 point) 13.1 degrees 17.3 degrees 12.7 degrees 14.5 degrees

Answer 1

In a right triangle, the angle opposite the longest side (hypotenuse) is 90 degrees. Here we have:

- \( a = 9 \) (the side opposite angle \( A \))
- \( b = 40 \)
- \( c = 41 \) (the hypotenuse)

Using the Law of Cosines:

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

We know \( c = 41 \) is the hypotenuse, so we'll set \( A = 90^\circ \). However, let's verify by plugging the values into the equation:

Plugging in the values:

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

Calculating the squares:

\[
81 = 1600 + 1681 - 3280 \cdot \cos A
\]

This simplifies to:

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

Now, rearranging for \( \cos A \):

\[
3280 \cdot \cos A = 3281 - 81
\]
\[
3280 \cdot \cos A = 3200
\]
\[
\cos A = \frac{3200}{3280}
\]
\[
\cos A = \frac{320}{328} = \frac{80}{82} \approx 0.9756097561
\]

Now to find angle \( A \):

\[
A = \cos^{-1}(0.9756097561)
\]

Using a calculator to find the angle:

\[
A \approx 12.7^\circ
\]

Thus, the value of angle \( A \) is:

\[
\boxed{12.7 \text{ degrees}}
\]