Question

Use the Law of Cosines,  b2=a2+c2−2ac⋅cosB , to find the value of angle B in a right triangle, where a=9 , b=40 , and c=41 .(1 point)Responses77.32 degrees77.32 degrees90.01 degrees90.01 degrees78.01 degrees78.01 degrees84.55 degrees84.55 degrees

Answer 1

In a right triangle, one of the angles is \(90^\circ\). We can still use the Law of Cosines to find the values of the angles. In your case, you want to find angle \(B\) with the given sides \(a = 9\), \(b = 40\), and \(c = 41\).

According to the Law of Cosines:

\[
b^2 = a^2 + c^2 - 2ac \cdot \cos B
\]

Plugging in the values:

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

Calculating the squares:

\[
1600 = 81 + 1681 - 738 \cdot \cos B
\]

Now combine:

\[
1600 = 1762 - 738 \cdot \cos B
\]

Rearranging gives:

\[
738 \cdot \cos B = 1762 - 1600
\]

Simplifying this:

\[
738 \cdot \cos B = 162
\]

Now solve for \(\cos B\):

\[
\cos B = \frac{162}{738} \approx 0.2195
\]

Now, take the inverse cosine to find angle \(B\):

\[
B = \cos^{-1}(0.2195) \approx 77.32^\circ
\]

Thus, the angle \(B\) is approximately \(77.32^\circ\).

The correct answer is:
**77.32 degrees**.