To find the length of the altitude \( h \) from vertex \( A \) to side \( BC \) in triangle \( ABC \), we can use the formula for the area of the triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we can take side \( a \) (which is opposite angle \( A \)) as the base.
Using the formula for the area involving two sides and the sine of the included angle, we can express the area of triangle \( ABC \) as:
\[ \text{Area} = \frac{1}{2} ab \sin(C) \]
In this case, since we only have side \( a \) and angle \( C \), we need to set up the relationship for the height:
\[ h = b \sin(C) \]
However, without knowing \( b \) or the side corresponding to \( A \), we will need to derive \( h \) a little differently. We can use the sine of angle \( C \) to relate \( h \) to side \( a = 36 , \text{cm} \).
First, let's find \( h \) using the angle \( C \):
\[ h = a \sin(C) = 36 \cdot \sin(23^\circ) \]
Calculating \( \sin(23^\circ) \):
\[ \sin(23^\circ) \approx 0.3907 \]
Now substituting back to find \( h \):
\[ h \approx 36 \cdot 0.3907 \approx 14.09 \]
Rounding this to the nearest whole number, we get:
\[ h \approx 14 , \text{cm} \]
The correct answer is 14.
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