In ΔVWX, *x* = 260 cm, m*m*∠V=80° and m*m*∠W=70°. Find the length of *w, to the nearest 10th of a centimeter.*

In the given triangle \( \Delta VWX \), we are provided with the following information:

- \( x = 260 \, \text{cm} \) (side opposite to angle \( W \)),
- \( m\angle V = 80^\circ \),
- \( m\angle W = 70^\circ \).

To find the length of side \( w \) (the side opposite to angle \( V \)), we can use the **Law of Sines**. The Law of Sines states that:

\[
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
\]

First, we calculate the remaining angle \( \angle X \):

\[
m\angle X = 180^\circ - m\angle V - m\angle W = 180^\circ - 80^\circ - 70^\circ = 30^\circ.
\]

Now, using the Law of Sines, we set up the following ratio:

\[
\frac{w}{\sin 80^\circ} = \frac{x}{\sin 30^\circ}.
\]

Substitute the known values:

\[
\frac{w}{\sin 80^\circ} = \frac{260}{\sin 30^\circ}.
\]

We know that \( \sin 80^\circ \approx 0.9848 \) and \( \sin 30^\circ = 0.5 \). So the equation becomes:

\[
\frac{w}{0.9848} = \frac{260}{0.5}.
\]

Now, solve for \( w \):

\[
w = \frac{260 \times 0.9848}{0.5} = 511.296.
\]

Thus, the length of side \( w \) is approximately **511.3 cm** (to the nearest 10th of a centimeter).

---

Would you like further details or clarifications on any step?

Here are 5 related questions to expand your understanding:
1. How does the Law of Sines apply when solving non-right triangles?
2. What is the significance of the angles in determining triangle properties?
3. Can you apply the Law of Cosines to verify this result?
4. How would you find the area of \( \Delta VWX \) using these dimensions?
5. How do changes in one angle affect the sides of a triangle?

**Tip:** The Law of Sines is especially useful when two angles and a side (AAS or ASA) are known in a triangle.

In ΔVWX, x = 260 cm, m∠V = 80° and m∠W = 70°. Find the length of w, to the nearest 10th of a centimeter.

Learn how to calculate the length of side w in triangle VWX, where x = 260 cm, m∠V = 80°, and m∠W = 70°. Using the Law of Sines and trigonometric ratios, this problem provides a step-by-step solution to finding w, approximately 511.3 cm. Suitable for students in Grades 9-11.

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