Find the measure of angle E in a triangle given the lengths of its sides.

1. Identify the problem type: This is a geometry problem involving a triangle where we need to find an angle given the lengths of all three sides. This suggests using the Law of Cosines.

1. Recall the Law of Cosines: The Law of Cosines states that for any triangle with sides $a$, $b$, and $c$, and the angle $C$ opposite side $c$, the following relationship holds: $c^2 = a^2 + b^2 - 2ab \cos(C)$. We can rearrange this formula to solve for the angle: $\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$.

3. Assign variables to the triangle: In triangle $GEF$, let:
- Side $GE = f = 8$
- Side $EF = g = 10$
- Side $GF = e = 17$
We want to find the measure of angle $E$, which is opposite side $GF$. So, in the Law of Cosines formula, we will use $E$ as our angle, $f$ and $g$ as the adjacent sides, and $e$ as the opposite side.

4. Apply the Law of Cosines to find angle E: Using the rearranged formula for the angle:
$$\cos(E) = \frac{f^2 + g^2 - e^2}{2fg}$$
Substitute the given side lengths:
$$\cos(E) = \frac{8^2 + 10^2 - 17^2}{2 \cdot 8 \cdot 10}$$

5. Calculate the values in the equation:
$$\cos(E) = \frac{64 + 100 - 289}{160}$$
$$\cos(E) = \frac{164 - 289}{160}$$
$$\cos(E) = \frac{-125}{160}$$

6. Simplify the fraction:
$$\cos(E) = -\frac{125}{160}$$
Divide both numerator and denominator by 5:
$$\cos(E) = -\frac{25}{32}$$

7. Find the angle E by taking the inverse cosine (arccos):
$$E = \arccos\left(-\frac{25}{32}\right)$$
Using a calculator:
$$E \approx 140.53556^\circ$$

1. Round the answer to the nearest tenth: The problem asks for the answer as an integer or a decimal rounded to the nearest tenth. Rounding $140.53556^\circ$ to the nearest tenth gives $140.5^\circ$.