Question

in triangle ABC, find the length, to the nearest tenth of an inch, of side a if b = 7 in., c = 10 in., and the angle formed by these sides measures 72 degrees.

Answer 1

To find the length of side \( a \) in triangle \( ABC \), we can use the Law of Cosines, which states:

\[
a^2 = b^2 + c^2 - 2bc \cdot \cos(A)
\]

where \( A \) is the angle opposite side \( a \), and \( b \) and \( c \) are the lengths of the other two sides.

In this case:
- \( b = 7 \) inches
- \( c = 10 \) inches
- \( A = 72^\circ \)

First, we need to calculate \( \cos(72^\circ) \). The value is approximately:

\[
\cos(72^\circ) \approx 0.309
\]

Now we can substitute the values into the Law of Cosines formula:

\[
a^2 = 7^2 + 10^2 - 2 \cdot 7 \cdot 10 \cdot \cos(72^\circ)
\]

Calculating \( 7^2 \) and \( 10^2 \):

\[
7^2 = 49
\]
\[
10^2 = 100
\]

Now substituting these values into the equation:

\[
a^2 = 49 + 100 - 2 \cdot 7 \cdot 10 \cdot 0.309
\]

Calculating \( 2 \cdot 7 \cdot 10 \cdot 0.309 \):

\[
2 \cdot 7 \cdot 10 \cdot 0.309 = 140 \cdot 0.309 = 43.26
\]

Now substituting back into the equation:

\[
a^2 = 49 + 100 - 43.26
\]
\[
a^2 = 149 - 43.26 = 105.74
\]

Now take the square root to find \( a \):

\[
a = \sqrt{105.74} \approx 10.3
\]

Thus, the length of side \( a \) is approximately **10.3 inches** to the nearest tenth.