Question

In ΔABC, 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.(1 point)
a=
in.

Answer 1

To find the length of side \( a \) in triangle \( \Delta ABC \) using the Law of Cosines, we can apply the formula:

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

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

First, we need to calculate \( \cos(72^\circ) \). You may use a calculator for this:

\[
\cos(72^\circ) \approx 0.3090
\]

Now, substitute the values into the formula:

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

Calculating the squares first:

\[
7^2 = 49
\]

\[
10^2 = 100
\]

Now plug in these values:

\[
a^2 = 49 + 100 - 2 \cdot 7 \cdot 10 \cdot 0.3090
\]

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

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

Now, we multiply this by \( \cos(72^\circ) \):

\[
140 \cdot 0.3090 \approx 43.26
\]

Now substitute this into the equation for \( a^2 \):

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

Now calculate:

\[
a^2 = 149 - 43.26 \approx 105.74
\]

Now, find \( a \) by taking the square root:

\[
a \approx \sqrt{105.74} \approx 10.3 \text{ inches}
\]

Thus, the length of side \( a \) to the nearest tenth of an inch is:

\[
\boxed{10.3} \text{ in.}
\]