Question

Use the stated variant of the Law of Cosines, a2=b2+c2−2bc⋅cosA 𝑎 2 = 𝑏 2 + 𝑐 2 − 2 𝑏 𝑐 ⋅ cos ⁡ 𝐴 , to find the value of angle A 𝐴 in a right triangle, where a=8 𝑎 = 8 , b=15 𝑏 = 15 , and c=17 𝑐 = 17 . Round your answer to the nearest whole number.(1 point)

Answer 1

To find the angle \( A \) in the triangle using the Law of Cosines, we start with the given values:

- \( a = 8 \)
- \( b = 15 \)
- \( c = 17 \)

The Law of Cosines in the form you provided is:

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

We can rearrange this equation to solve for \( \cos A \):

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

Now, we substitute the values of \( a \), \( b \), and \( c \) into the equation:

1. Calculate \( b^2 + c^2 - a^2 \):

\[
b^2 = 15^2 = 225
\]
\[
c^2 = 17^2 = 289
\]
\[
a^2 = 8^2 = 64
\]

So,

\[
b^2 + c^2 - a^2 = 225 + 289 - 64 = 450
\]

2. Calculate \( 2bc \):

\[
2bc = 2 \cdot 15 \cdot 17 = 510
\]

3. Now plug these values into the cosine equation:

\[
\cos A = \frac{450}{510} = \frac{15}{17}
\]

4. To find \( A \), we take the arccosine:

\[
A = \arccos\left(\frac{15}{17}\right)
\]

Now, we calculate \( A \):

Using a calculator (make sure it's in degree mode),

\[
A \approx \arccos\left(\frac{15}{17}\right) \approx 42.51^\circ
\]

Finally, rounding to the nearest whole number, we find:

\[
\boxed{43}
\]