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

1. Identify the problem type: This is a geometry problem that can be solved using the Law of Cosines, as we are given the lengths of all three sides of a triangle and asked to find the measure of one of its angles.

1. State the Law of Cosines: The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. For a triangle with sides $a$, $b$, and $c$, and the angle $C$ opposite side $c$, the formula is $c^2 = a^2 + b^2 - 2ab \cos(C)$. To find an angle, we can rearrange this formula to $\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$.

3. Assign variables to the triangle sides and angle: In the given triangle, we want to find the measure of angle $V$. Let's denote the side opposite angle $V$ as $v$, the side opposite angle $W$ as $w$, and the side opposite angle $X$ as $x$. From the diagram, we have:
- Side $x$ (opposite angle $X$) = 13
- Side $w$ (opposite angle $W$) = 5
- Side $v$ (opposite angle $V$) = 9

We want to find $m\angle V$. So, we will use the Law of Cosines in the form:
$$v^2 = w^2 + x^2 - 2wx \cos(V)$$

4. Rearrange the formula to solve for $\cos(V)$:
$$2wx \cos(V) = w^2 + x^2 - v^2$$
$$\cos(V) = \frac{w^2 + x^2 - v^2}{2wx}$$

5. Substitute the given side lengths into the formula:
$$w = 5$$ $$x = 13$$ $$v = 9$$
$$\cos(V) = \frac{5^2 + 13^2 - 9^2}{2 \cdot 5 \cdot 13}$$

6. Calculate the squares of the side lengths:
$$5^2 = 25$$ $$13^2 = 169$$ $$9^2 = 81$$ $$\cos(V) = \frac{25 + 169 - 81}{2 \cdot 5 \cdot 13}$$

7. Perform the addition and subtraction in the numerator:
$$25 + 169 = 194$$ $$194 - 81 = 113$$ $$\cos(V) = \frac{113}{2 \cdot 5 \cdot 13}$$

8. Calculate the product in the denominator:
$$2 \cdot 5 \cdot 13 = 10 \cdot 13 = 130$$ $$\cos(V) = \frac{113}{130}$$

9. Calculate the value of $\cos(V)$:
$$\cos(V) \approx 0.869230769$$

10. Find the measure of angle $V$ by taking the inverse cosine (arccosine):
$$m\angle V = \arccos\left(\frac{113}{130}\right)$$
$$m\angle V \approx \arccos(0.869230769)$$

11. Calculate the angle in degrees and round to the nearest tenth:
Using a calculator, $\arccos(0.869230769) \approx 29.6155^\circ$.

Correction: Let's re-evaluate the assignment of sides. The angle we want to find is $V$. The side opposite to $V$ is $W X$, which has length 13. The side opposite to $W$ is $V X$, which has length 5. The side opposite to $X$ is $V W$, which has length 9. So, let $v=13$, $w=5$, $x=9$. We want to find $m\angle V$. The Law of Cosines states: $v^2 = w^2 + x^2 - 2wx \cos(V)$. Rearranging for $\cos(V)$: $\cos(V) = \frac{w^2 + x^2 - v^2}{2wx}$. Substituting the values: $\cos(V) = \frac{5^2 + 9^2 - 13^2}{2 \cdot 5 \cdot 9}$.

12. Recalculate with correct side assignments:
$$w = 5$$ $$x = 9$$ $$v = 13$$
$$\cos(V) = \frac{5^2 + 9^2 - 13^2}{2 \cdot 5 \cdot 9}$$
$$\cos(V) = \frac{25 + 81 - 169}{90}$$
$$\cos(V) = \frac{106 - 169}{90}$$
$$\cos(V) = \frac{-63}{90}$$
$$\cos(V) = -\frac{7}{10} = -0.7$$

13. Find the measure of angle $V$ by taking the inverse cosine:
$$m\angle V = \arccos(-0.7)$$

14. Calculate the angle in degrees and round to the nearest tenth:
Using a calculator, $\arccos(-0.7) \approx 134.427^\circ$. Rounding to the nearest tenth, we get $134.4^\circ$.

Self-correction: The diagram labels the vertices X, V, W. The side lengths are given as: XV = 5, VW = 9, XW = 13. We need to find $m\angle V$. Let's use the Law of Cosines with the angle $V$ as the target. The side opposite to angle $V$ is $XW$, so $v = 13$. The other two sides are $XV$ and $VW$. Let $x = 5$ (side opposite to $X$) and $w = 9$ (side opposite to $W$). The Law of Cosines is $v^2 = x^2 + w^2 - 2xw \cos(V)$. Rearranging to solve for $\cos(V)$: $\cos(V) = \frac{x^2 + w^2 - v^2}{2xw}$.

15. Substitute the correct values into the formula:
$$x = 5$$ $$w = 9$$ $$v = 13$$
$$\cos(V) = \frac{5^2 + 9^2 - 13^2}{2 \cdot 5 \cdot 9}$$

16. Calculate the squares and perform the operations:
$$5^2 = 25$$ $$9^2 = 81$$ $$13^2 = 169$$
$$\cos(V) = \frac{25 + 81 - 169}{2 \cdot 5 \cdot 9}$$
$$\cos(V) = \frac{106 - 169}{90}$$
$$\cos(V) = \frac{-63}{90}$$
$$\cos(V) = -0.7$$

17. Find the angle $V$ using the inverse cosine function:
$$m\angle V = \arccos(-0.7)$$

18. Calculate the final answer and round:
$$m\angle V \approx 134.427^\circ$$
Rounding to the nearest tenth, $m\angle V \approx 134.4^\circ$.

Final check of side assignments based on diagram: The vertices are X, V, W. The side lengths are: XV = 5, VW = 9, XW = 13. We want to find $m\angle V$. The side opposite to $\angle V$ is $XW$, so its length is 13. The side opposite to $\angle X$ is $VW$, so its length is 9. The side opposite to $\angle W$ is $XV$, so its length is 5. Let $a=5$, $b=9$, $c=13$. We want to find the angle opposite side $c$, which is $\angle V$. The Law of Cosines is $c^2 = a^2 + b^2 - 2ab \cos(C)$. In our case, $13^2 = 5^2 + 9^2 - 2(5)(9) \cos(V)$.

19. Apply the Law of Cosines with correct vertex-side correspondence:
Let $a = XV...