Question

in \\( \triangle stu, s = 8.5 \mathrm { cm }, t = 4.4 \mathrm { cm } \\) and \\( \angle u = 59 ^ { \circ } \\). find the length of \\( u \\), to the nearest 10th of a centimeter.

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines formula is \(u^{2}=s^{2}+t^{2}-2st\cos U\).
Substitute \(s = 8.5\), \(t = 4.4\), and \(U = 59^{\circ}\) into the formula:
\(u^{2}=8.5^{2}+4.4^{2}-2\times8.5\times4.4\times\cos(59^{\circ})\)

Step2: Calculate each term

First, calculate \(8.5^{2}=72.25\), \(4.4^{2}=19.36\).
Then, calculate \(2\times8.5\times4.4 = 74.8\), and \(\cos(59^{\circ})\approx0.5150\).
So \(74.8\times0.5150 = 38.522\)

Step3: Compute \(u^{2}\)

\(u^{2}=72.25 + 19.36-38.522\)
\(u^{2}=91.61 - 38.522\)
\(u^{2}=53.088\)

Step4: Find \(u\)

Take the square - root of \(u^{2}\): \(u=\sqrt{53.088}\approx7.3\)

Answer:

\(7.3\)