Question

fall 2025 geometry b wwva law of cosines dallas 720 mi destination 1 125° 290 mi destination 2 dropping off his cargo, he flies southeast 290 miles to his second destination. if the angle formed by his trip is 125°, what is the distance he will fly from the second destination back to dallas? 842,026 miles 490 miles 918 miles

Explanation:

Step1: Identify the Law of Cosines formula

The Law of Cosines for a triangle with sides \(a\), \(b\), and \(c\), and the angle \(C\) opposite side \(c\) is \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\). In this problem, let \(a = 720\) miles, \(b=290\) miles, and \(C = 125^{\circ}\), and we need to find the length of the side opposite the \(125^{\circ}\) angle (the distance from Destination 2 to Dallas), let's call it \(c\).

Step2: Substitute the values into the formula

First, calculate \(a^{2}\), \(b^{2}\), and \(2ab\cos(C)\):

• \(a^{2}=720^{2}=518400\)
• \(b^{2}=290^{2} = 84100\)
• \(2ab=2\times720\times290=2\times208800 = 417600\)
• \(\cos(125^{\circ})\approx\cos(180^{\circ} - 55^{\circ})=-\cos(55^{\circ})\approx - 0.5736\)
• Then \(2ab\cos(C)=417600\times(- 0.5736)\approx - 417600\times0.5736=-239535.36\)

Now, substitute into the Law of Cosines formula:
\(c^{2}=a^{2}+b^{2}-2ab\cos(C)=518400 + 84100-(-239535.36)=518400 + 84100+239535.36\)
\(c^{2}=518400+84100 = 602500\); \(602500+239535.36 = 842035.36\) (approximate, slight difference due to \(\cos\) approximation)

Step3: Take the square root of \(c^{2}\)

\(c=\sqrt{842035.36}\approx918\) miles (since \(\sqrt{842026}\approx918\) as well, considering possible rounding in the problem)

Answer:

918 miles (corresponding to the option "918 miles")