Question

in δtuv, t = 670 cm, u = 510 cm and ∠v = 159°. find the length of v, to the nearest centimeter.

Explanation:

Step1: Recall the Law of Cosines

The Law of Cosines states that for a triangle with sides \(a\), \(b\), \(c\) and the angle \(C\) opposite side \(c\), \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\). In \(\triangle TUV\), we want to find the length of \(v\), and we know \(t = 670\) cm, \(u=510\) cm, and \(\angle V = 159^{\circ}\). So, by the Law of Cosines, \(v^{2}=t^{2}+u^{2}-2tu\cos(V)\).

Step2: Substitute the given values

Substitute \(t = 670\), \(u = 510\), and \(V=159^{\circ}\) into the formula. First, calculate \(t^{2}=670^{2}=448900\), \(u^{2}=510^{2}=260100\), and \(2tu=2\times670\times510 = 2\times341700=683400\). Then, \(\cos(159^{\circ})\approx\cos(180^{\circ} - 21^{\circ})=-\cos(21^{\circ})\approx - 0.9336\).

So, \(v^{2}=448900 + 260100-2\times670\times510\times\cos(159^{\circ})\)
\(v^{2}=709000-683400\times(- 0.9336)\)
\(v^{2}=709000 + 683400\times0.9336\)

Calculate \(683400\times0.9336\approx683400\times0.9336 = 638022.24\)

Then \(v^{2}=709000+638022.24 = 1347022.24\)

Step3: Take the square root

Take the square root of \(v^{2}\) to find \(v\). \(v=\sqrt{1347022.24}\approx1160\) (rounded to the nearest centimeter)

Answer:

\(1160\)