Question

in triangle ABC, find the length, to the nearest tenth of an inch, of side a if b = 8 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 \( ABC \) using the Law of Cosines, we can use the following formula:

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

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

In your case, we have:

- \( b = 8 \) inches
- \( c = 10 \) inches
- \( A = 72^\circ \)

First, we need to calculate \( a^2 \):

\[
a^2 = 8^2 + 10^2 - 2 \cdot 8 \cdot 10 \cdot \cos(72^\circ)
\]

Calculating the values:

\[
8^2 = 64
\]
\[
10^2 = 100
\]
So,

\[
a^2 = 64 + 100 - 2 \cdot 8 \cdot 10 \cdot \cos(72^\circ)
\]

Next, let's calculate \( \cos(72^\circ) \). Using a calculator:

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

Now substituting this value into the equation:

\[
a^2 = 64 + 100 - 2 \cdot 8 \cdot 10 \cdot 0.309
\]
\[
= 164 - 2 \cdot 8 \cdot 10 \cdot 0.309
\]
\[
= 164 - 2 \cdot 80 \cdot 0.309
\]
\[
= 164 - 49.44
\]
\[
= 114.56
\]

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

\[
a = \sqrt{114.56} \approx 10.7
\]

Thus, the length of side \( a \) is approximately \( \boxed{10.7} \) inches, rounded to the nearest tenth of an inch.