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:

\[ \boxed{43} \]