Question

in \\( \delta wxy, w = 13 \mathrm { cm }, x = 12 \mathrm { cm } \\) and \\( \angle y = 18 ^ { \circ } \\). find the length of \\( y \\), to the nearest centimeter.

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines formula is \(y^{2}=w^{2}+x^{2}-2wx\cos Y\).
Substitute \(w = 13\), \(x = 12\), and \(Y=18^{\circ}\) (so \(\cos Y=\cos18^{\circ}\approx0.9511\)) into the formula:
\(y^{2}=13^{2}+12^{2}-2\times13\times12\times0.9511\)

Step2: Calculate each term

First, \(13^{2}=169\), \(12^{2}=144\), and \(2\times13\times12\times0.9511 = 304.6176\)
Then \(y^{2}=169 + 144-304.6176\)
\(y^{2}=313-304.6176\)
\(y^{2}=8.3824\)

Step3: Solve for \(y\)

Take the square root of both sides: \(y=\sqrt{8.3824}\approx 3\)

Answer:

\(4\)