Certainly! Let's solve for the angle [tex]\( C \)[/tex] using the Law of Cosines:
The Law of Cosines states:
[tex]\[ c^2 = a^2 + b^2 - 2ab \cdot \cos C \][/tex]
We need to solve for [tex]\( C \)[/tex]. First, we can rearrange the formula to isolate [tex]\( \cos C \)[/tex]:
[tex]\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \][/tex]
Now, let's plug in the given values [tex]\( a = 7 \)[/tex], [tex]\( b = 10 \)[/tex], and [tex]\( c = 5 \)[/tex]:
[tex]\[ \cos C = \frac{7^2 + 10^2 - 5^2}{2 \cdot 7 \cdot 10} \][/tex]
Calculating the squares and the product in the denominator:
[tex]\[ \cos C = \frac{49 + 100 - 25}{140} \][/tex]
[tex]\[ \cos C = \frac{124}{140} \][/tex]
[tex]\[ \cos C = 0.8857142857142857 \][/tex]
Next, we need to find [tex]\( C \)[/tex] by taking the arccosine (inverse cosine) of the calculated value:
[tex]\[ C = \arccos(0.8857142857142857) \][/tex]
This gives us [tex]\( C \)[/tex] in radians:
[tex]\[ C \approx 0.48276592332573415 \text{ radians} \][/tex]
To convert radians to degrees:
[tex]\[ C_{\text{degrees}} = C \times \frac{180}{\pi} \][/tex]
Substituting the value we have:
[tex]\[ C_{\text{degrees}} \approx 0.48276592332573415 \times 57.2958 \][/tex]
[tex]\[ C_{\text{degrees}} \approx 27.660449899300872 \][/tex]
Therefore, the cosine of angle [tex]\( C \)[/tex] is approximately [tex]\( 0.8857142857142857 \)[/tex], the angle [tex]\( C \)[/tex] in radians is approximately [tex]\( 0.48276592332573415 \)[/tex] radians, and the angle [tex]\( C \)[/tex] in degrees is approximately [tex]\( 27.66^\circ \)[/tex].
