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

8 meters
8 meters

13 meters
13 meters

6 meters
6 meters

16 meters

Answer 1

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:

1. As \( A = \frac{1}{2} a \cdot h \)
2. 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 \( a \):

1. 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**.