Answer:
1348 light years.
Step-by-step explanation:
Please find the attachment.
Let x represent the distance between KA-7 and KA-11.
We have been given that the first system, KA-7, is 1200 light years away while the second system, KA-11, is 1700 light years away. Lucy sees an angle of 52 degrees between KA-7 and KA-11.
We can see from our attachment that Lucy, KA-7 and KA-11 forms a triangle and we will use law of cosines to solve for x.
[tex](AB)^2=(AC)^2+(BC)^2-2(AC)(BC)*cos (C)[/tex]
Upon substituting our given values in above formula we will get,
[tex]x^2=1200^2+1700^2-2*1200*1700*cos (52^o)[/tex]
[tex]x^2=1440000+2890000-4080000*0.615661475[/tex]
[tex]x^2=4330000-2511898.81933008[/tex]
[tex]x^2=1818101.18066992[/tex]
Let us take square root of both sides of our equation.
[tex]x=\sqrt{1818101.18066992}[/tex]
[tex]x=1348.3698234\approx 1348[/tex]
Therefore, KA-7 and KA-11 are approximately 1348 light years apart.
The distance (straight line distance) between the star systems KA-7 and KA-11 is 1348.37
Let there is triangle ABC such that |AB| = a units, |AC| = b units, and |BC| = c units and the internal angle A is of θ degrees, then we have:
[tex]a^2 + b^2 -2ab\cos(\theta) = c^2[/tex]
(c is opposite side to angle A)
We can use law of cosines here because, as shown in the figure below, we have lengths of two sides of a triangle and the angle between it and we have to find the side length opposite to the known angle.
We've got:
|AB| = a = 1200 light years
|AC| = b = 1700 light years
|BC| = c = distance to be known
∠BAC = 52°
Then, using the cosine law gives:
[tex]c^2 =a^2 + b^2 -2ab\cos(\theta)\\\\c =\sqrt{a^2 + b^2 -2ab\cos(\theta) }\\\\\\c= \sqrt{1200^2 + 1700^2 -2(1200)(1700)\cos(52^\circ)}\\\\c \approx \sqrt{433000 -2511898.82} \approx 1348.37[/tex](in light years).
Thus, the distance (straight line distance) between the star systems KA-7 and KA-11 is 1348.37
Learn more about law of cosines here:
https://brainly.com/question/17289163
