Question

In ΔUVW, u = 1 inches, v = 5.3 inches and ∠W=151°. Find the length of w, to the nearest 10th of an inch.
Answer
Attempt 1 out of 2

Answer 1

To find the length of side \( w \) in triangle \( \Delta UVW \), where you know the lengths of sides \( u \) and \( v \) and the included angle \( \angle W \), you can use the Law of Cosines. This law states:

\[
w^2 = u^2 + v^2 - 2uv \cdot \cos(\angle W)
\]

In your case, \( u = 1 \) inch, \( v = 5.3 \) inches, and \( \angle W = 151^\circ \).

First, calculate \( \cos(151^\circ) \):

Using a calculator:
\[
\cos(151^\circ) \approx -0.9063
\]

Now substitute the values into the Law of Cosines formula:

\[
w^2 = (1)^2 + (5.3)^2 - 2 \cdot (1) \cdot (5.3) \cdot (-0.9063)
\]

Calculate each term:

1. \( (1)^2 = 1 \)
2. \( (5.3)^2 \approx 28.09 \)
3. \( -2 \cdot (1) \cdot (5.3) \cdot (-0.9063) = 2 \cdot (5.3) \cdot (0.9063) \approx 9.604 \)

Now combine these values:

\[
w^2 = 1 + 28.09 + 9.604 = 38.694
\]

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

\[
w \approx \sqrt{38.694} \approx 6.2 \, \text{inches}
\]

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

\[
\boxed{6.2} \, \text{inches}
\]