Question

question in δtuv, t = 150 inches, u = 580 inches and ∠v = 105°. find the length of v, to the nearest inch. score: 0/3 penalty: none law of cosines (sas) answer attempt 1 out of 2 submit answer watch video show examples

Explanation:

Step1: Recall the Law of Cosines formula

The Law of Cosines for finding side \(v\) in \(\triangle TUV\) is \(v^{2}=t^{2}+u^{2}-2tu\cos V\). Given \(t = 150\), \(u = 580\), and \(V=105^{\circ}\), and \(\cos(105^{\circ})=\cos(60^{\circ}+ 45^{\circ})=\cos60^{\circ}\cos45^{\circ}-\sin60^{\circ}\sin45^{\circ}=\frac{1}{2}\times\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{2}-\sqrt{6}}{4}\approx - 0.2588\).

Step2: Substitute values into the formula

\[

$$\begin{align*} v^{2}&=150^{2}+580^{2}-2\times150\times580\times\cos(105^{\circ})\\ v^{2}&=22500 + 336400-174000\times(- 0.2588)\\ v^{2}&=22500+336400 + 44931.2\\ v^{2}&=403831.2 \end{align*}$$

\]

Step3: Solve for \(v\)

\(v=\sqrt{403831.2}\approx635\)

Answer:

635