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)

Responses

12.7 degrees
12.7 degrees

17.3 degrees
17.3 degrees

13.1 degrees
13.1 degrees

14.5 degrees

1 answer

To find the value of angle \( A \) in a triangle using the Law of Cosines, we can rearrange the formula \( a^2 = b^2 + c^2 - 2bc \cdot \cos A \) to solve for \( \cos A \):

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

Given the values:

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

We can first compute \( b^2 \), \( c^2 \), and \( a^2 \):

\[ b^2 = 40^2 = 1600 \] \[ c^2 = 41^2 = 1681 \] \[ a^2 = 9^2 = 81 \]

Now, we substitute these values into the cosine formula:

\[ \cos A = \frac{1600 + 1681 - 81}{2 \cdot 40 \cdot 41} \] \[ \cos A = \frac{1600 + 1681 - 81}{3280} \] \[ \cos A = \frac{3200}{3280} \] \[ \cos A = \frac{320}{328} \approx 0.9756 \]

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

\[ A = \cos^{-1}(0.9756) \]

Using a calculator:

\[ A \approx 12.7^\circ \]

Therefore, the value of angle \( A \) is 12.7 degrees.