Question

In ΔWXY, w = 8 inches, x = 2.3 inches and ∠Y=136°. Find the length of y, to the nearest 10th of an inch.

Answer 1

To find the length \( y \) in triangle \( \Delta WXY \) with given sides \( w \), \( x \), and angle \( \angle Y \), we can use the Law of Cosines, which states:

\[
y^2 = w^2 + x^2 - 2wx \cos(Y)
\]

Given:
- \( w = 8 \) inches
- \( x = 2.3 \) inches
- \( \angle Y = 136^\circ \)

First, we need to calculate \( \cos(136^\circ) \):

\[
\cos(136^\circ) = \cos(180^\circ - 44^\circ) = -\cos(44^\circ) \approx -0.7193
\]

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

\[
y^2 = 8^2 + 2.3^2 - 2 \cdot 8 \cdot 2.3 \cdot (-0.7193)
\]

Calculating each term:

\[
8^2 = 64
\]
\[
2.3^2 \approx 5.29
\]
\[
-2 \cdot 8 \cdot 2.3 \cdot (-0.7193) = 2 \cdot 8 \cdot 2.3 \cdot 0.7193
\]

Calculating \( 2 \cdot 8 \cdot 2.3 \):

\[
2 \cdot 8 = 16
\]
\[
16 \cdot 2.3 = 36.8
\]

Now we find:

\[
36.8 \cdot 0.7193 \approx 26.46984
\]

Putting it all together:

\[
y^2 = 64 + 5.29 + 26.46984
\]

Calculating the sum:

\[
y^2 \approx 64 + 5.29 + 26.46984 \approx 95.75984
\]

Now, taking the square root to find \( y \):

\[
y \approx \sqrt{95.75984} \approx 9.8 \text{ inches}
\]

Therefore, the length of \( y \) to the nearest tenth of an inch is:

\[
\boxed{9.8} \text{ inches}
\]