Question

a triangle upper A upper B upper C
If h=10
meters and m∠C = 39°
, then what is the length of side a
? Round your answer to the nearest whole number.

(1 point)
Responses

13 meters
13 meters

8 meters
8 meters

16 meters
16 meters

6 meters

Answer 1

To find the length of side \( a \) in triangle \( ABC \) using the given height \( h = 10 \) meters and angle \( C = 39^\circ \), we can use the formula for the area of a triangle \( A \) in two different ways.

1. **Using height and base**:
The area \( A \) of triangle \( ABC \) can be calculated using the height to side \( a \):
\[
A = \frac{1}{2} \times a \times h
\]
Substituting for \( h \):
\[
A = \frac{1}{2} \times a \times 10 = 5a
\]

2. **Using two sides and the sine of angle C**:
Another way to find the area is:
\[
A = \frac{1}{2} \times b \times c \times \sin(C)
\]
However, we need side \( a \) and don't currently have \( b \) and \( c \).

3. Since we are given the height \( h \) and angle \( C \), we can also relate the height to side \( a \):
The height \( h \) from vertex \( C \) splits \( a \) into two segments \( a_1 \) and \( a_2 \) where:
\[
h = b \sin(C)
\]
Here, the opposite side to angle \( C \) also relates to base \( a \):
\[
a = h \cdot \cot(C)
\]

Therefore:
\[
a = 10 \cdot \cot(39^\circ)
\]

4. Calculating \( \cot(39^\circ) \):
\[
\cot(39^\circ) = \frac{1}{\tan(39^\circ)} \approx 1.202 (using a calculator)
\]
Thus,
\[
a \approx 10 \cdot 1.202 \approx 12.02
\]

5. Rounding to the nearest whole number, we can conclude that:
\[
a \approx 12 \text{ meters}
\]

It appears that based on the provided options, the closest option is 13 meters. Hence the final answer is:
**13 meters.**