To find the length of side \( a \) in triangle \( ABC \) with height \( h = 10 \) meters to side \( a \) and angle \( C = 39^\circ \), we can use the relationship between the height of the triangle, angle \( C \), and side \( a \).
In triangle \( ABC \), the height \( h \) can be expressed as:
\[ h = b \cdot \sin(C) \]
where \( b \) is the side opposite angle \( C \), which is side \( a \). However, we can also express the relations using basic properties of triangles. The height \( h \) is perpendicular from vertex \( C \) to side \( a \).
We also know that the area \( A \) of triangle \( ABC \) can be expressed in two ways:
- As \( A = \frac{1}{2} a \cdot h \)
- As \( A = \frac{1}{2} b \cdot c \cdot \sin(C) \)
Since \( h = 10 \) m and \( \sin(C) = \sin(39^\circ) \) can be calculated:
First, let's find \( \sin(39^\circ) \):
\[ \sin(39^\circ) \approx 0.6293 \]
Now we can express area in terms of side \( a \):
\[ A = \frac{1}{2} a \cdot 10 \]
This gives us:
\[ A = 5a \]
Now we can use the fact that:
\[ A = \frac{1}{2} \cdot a \cdot a \cdot \sin(39^\circ) \]
Thus,
\[ A = \frac{1}{2} \cdot 10 \cdot b \cdot \sin(39^\circ) \]
Equating the two expressions:
\[ 5a = \frac{1}{2} \cdot a \cdot 10 \cdot \sin(39^\circ) \]
We can simplify this if we divide both sides by \( \frac{1}{2} \) and isolate \( a \):
- Solve for \( a \):
\[ a = \frac{5 \cdot 10}{10 \cdot \sin(39^\circ)} \]
\[ a = \frac{5}{\sin(39^\circ)} \]
Calculate:
\[ \sin(39^\circ) \approx 0.6293 \]
Now substitute the values:
\[ a \approx \frac{5}{0.6293} \approx 7.95 \approx 8 \text{ meters} \]
Thus, rounding to the nearest whole number, the length of side \( a \) is approximately:
8 meters.
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