Question

in δefg, e = 830 cm, f = 630 cm and g = 600 cm. find the measure of ∠g to the nearest 10th of a degree. answer attempt 3 out of 3

Explanation:

Step1: Recall the Law of Cosines

For a triangle with sides \(a\), \(b\), \(c\) and the angle \(C\) opposite side \(c\), the Law of Cosines states \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\). In \(\triangle EFG\), \(\angle G\) is opposite side \(g\), so we have \(g^{2}=e^{2}+f^{2}-2ef\cos(G)\).

Step2: Rearrange the formula to solve for \(\cos(G)\)

\[

$$\begin{align*} 2ef\cos(G)&=e^{2}+f^{2}-g^{2}\\ \cos(G)&=\frac{e^{2}+f^{2}-g^{2}}{2ef} \end{align*}$$

\]

Step3: Substitute the given values

We know \(e = 830\), \(f = 630\), \(g = 600\). Substitute these into the formula:
\[

$$\begin{align*} \cos(G)&=\frac{830^{2}+630^{2}-600^{2}}{2\times830\times630}\\ &=\frac{688900 + 396900- 360000}{1045800}\\ &=\frac{725800}{1045800}\\ &\approx0.6939 \end{align*}$$

\]

Step4: Find the angle \(G\)

Take the inverse cosine of \(0.6939\) to find \(G\): \(G=\cos^{-1}(0.6939)\approx46.0^{\circ}\) (to the nearest tenth of a degree)

Answer:

\(46.0^{\circ}\)