Question

lucy’s spaceship is low on fuel. lucy is 1,200 light years (l.y.) from the star system ka-7, as shown. she is 1,700 l.y. from the system ka-11. lucy sees an angle of 52° between ka-7 and ka-11. to decide whether to visit ka-11, lucy needs to know the distance between the systems. how far apart are ka-7 and ka-11? do not round during your calculations. round your final answer to the nearest hundred light years. l.y.

Explanation:

Step1: Identify the formula

We can use the Law of Cosines to find the distance between KA - 7 and KA - 11. The Law of Cosines states that for a triangle with sides \(a\), \(b\), and \(c\) and the angle \(C\) opposite side \(c\), \(c^{2}=a^{2}+b^{2}-2ab\cos(C)\). Let \(a = 1200\), \(b = 1700\), and \(C=52^{\circ}\), and we want to find the length of the side opposite the \(52^{\circ}\) angle (let's call it \(c\)).

Step2: Substitute the values into the formula

Substitute \(a = 1200\), \(b = 1700\), and \(\cos(52^{\circ})\) into the formula:
\(c^{2}=1200^{2}+1700^{2}-2\times1200\times1700\times\cos(52^{\circ})\)
First, calculate \(1200^{2}=1440000\), \(1700^{2} = 2890000\), and \(2\times1200\times1700=4080000\)
Then, \(\cos(52^{\circ})\approx0.6157\) (using a calculator)
So, \(c^{2}=1440000 + 2890000-4080000\times0.6157\)
\(c^{2}=4330000-4080000\times0.6157\)
Calculate \(4080000\times0.6157 = 4080000\times0.6+4080000\times0.0157=2448000+64056 = 2512056\)
Then \(c^{2}=4330000 - 2512056=1817944\)

Step3: Take the square root

Take the square root of \(c^{2}\) to find \(c\):
\(c=\sqrt{1817944}\approx1348.31\)

Step4: Round to the nearest hundred

Rounding \(1348.31\) to the nearest hundred, we look at the tens digit (4). Since \(4<5\), we round down. So \(1348.31\approx1300\)

Answer:

1300