Question

Use the Law of Cosines, a2=b2+c2−2bc⋅cosA, to find the value of angle A in a right triangle, where a=9, b=40, and c=41.(1 point)

Answer 1

In a right triangle, we typically have one angle, say angle A, that is opposite side a. Given the values a = 9, b = 40, and c = 41, we can use the Law of Cosines:

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

Plugging in the known values, we have:

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

Calculating the squares gives:

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

Next, we simplify this equation:

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

Rearranging it to isolate the cosine term:

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

This simplifies to:

\[
3280 \cdot \cos A = 3200
\]

Now, dividing both sides by 3280:

\[
\cos A = \frac{3200}{3280}
\]

We can simplify this fraction:

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

Now, to find the angle A, we take the inverse cosine:

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

Using a calculator,

\[
A \approx 16.26^\circ
\]

Therefore, the value of angle A is approximately \( 16.26^\circ \).