Certainly! Let's solve for angle [tex]\( C \)[/tex] in the given triangle with sides [tex]\( a = 4 \)[/tex], [tex]\( b = 7 \)[/tex], and [tex]\( c = 9 \)[/tex].
First, we start from the given equation which is an application of the Law of Cosines:
[tex]\[
c^2 = a^2 + b^2 - 2ab \cos C
\][/tex]
Substitute the known side lengths into the equation:
[tex]\[
9^2 = 4^2 + 7^2 - 2 \cdot 4 \cdot 7 \cos C
\][/tex]
Evaluate the squares of the sides:
[tex]\[
81 = 16 + 49 - 56 \cos C
\][/tex]
Combine the terms on the right-hand side:
[tex]\[
81 = 65 - 56 \cos C
\][/tex]
Now isolate the term involving [tex]\( \cos C \)[/tex]:
[tex]\[
81 - 65 = -56 \cos C
\][/tex]
Simplify:
[tex]\[
16 = -56 \cos C
\][/tex]
Divide both sides by [tex]\(-56\)[/tex] to solve for [tex]\( \cos C \)[/tex]:
[tex]\[
\cos C = \frac{16}{-56} = -\frac{2}{7} \approx -0.2857142857142857
\][/tex]
Now to find the angle [tex]\( C \)[/tex], take the arccos (inverse cosine) of [tex]\(\cos C\)[/tex].
[tex]\[
C = \arccos\left(-\frac{2}{7}\right)
\][/tex]
This yields:
[tex]\[
C \approx 1.8605480282309441 \text{ radians}
\][/tex]
To convert this angle from radians to degrees, we use the conversion factor [tex]\( \frac{180}{\pi} \)[/tex]:
[tex]\[
C \approx 1.8605480282309441 \times \frac{180}{\pi} \approx 106.60154959902025^\circ
\][/tex]
So, the angle [tex]\( C \)[/tex] is approximately:
[tex]\[
C \approx 106.60^\circ
\][/tex]
Thus, the angle [tex]\( C \)[/tex] is about [tex]\( 106.60^\circ \)[/tex] when rounded to two decimal places.
