To find the length \( WY \) and the area of triangle \( XYZ \), we'll need to apply some trigonometry. However, the problem does not clearly define the positions or relationships between the points and the angles involved.
Assuming the following:
- \( XY = 5 \) m
- \( YZ = 6 \) m
- \( \angle XYZ = 67^\circ \)
Let's find \( WY \) (which seems to be the height from \( W \) to side \( XY \)) if \( W \) is at the top (opposite vertex).
1. Calculate the third side \( XZ \) using the Law of Cosines
\[ XZ^2 = XY^2 + YZ^2 - 2 \cdot XY \cdot YZ \cdot \cos(\angle XYZ) \]
Plugging in the values:
\[ XZ^2 = 5^2 + 6^2 - 2 \cdot 5 \cdot 6 \cdot \cos(67^\circ) \]
Calculating the values:
\[ XZ^2 = 25 + 36 - 60 \cdot \cos(67^\circ) \] \[ XZ^2 = 61 - 60 \cdot 0.3907 \quad (\text{cosine of } 67^\circ \text{ is approximately } 0.3907) \] \[ XZ^2 = 61 - 23.442 \approx 37.558 \] \[ XZ \approx \sqrt{37.558} \approx 6.13 \]
2. Calculate the area of triangle \( XYZ \)
The area \( A \) of triangle can be calculated using the formula:
\[ A = \frac{1}{2} \cdot base \cdot height \]
We can also use the formula based on two sides and the included angle:
\[ A = \frac{1}{2} \cdot XY \cdot YZ \cdot \sin(\angle XYZ) \]
Substituting our known values:
\[ A = \frac{1}{2} \cdot 5 \cdot 6 \cdot \sin(67^\circ) \]
Calculating \( \sin(67^\circ) \approx 0.9200 \):
\[ A = \frac{1}{2} \cdot 5 \cdot 6 \cdot 0.9200 \] \[ A = 15 \cdot 0.9200 \approx 13.8 \]
Summary
- The side length \( XZ \approx 6.13 \) m, but it's not needed because only WY was referred to as height (which we do not explicitly calculate here).
- The area of triangle \( XYZ \) is approximately \( 13.8 \) square meters.
So we have:
- \( WY \) unspecified, or you need to derive it from the geometric construction, and
- Area \( A \approx 13.8 \) m².