Question

satellites ka-12 and sal-1 have spotted a ufo. scientists want to determine its distance from ka-12 so they can later determine its size. the distance between these satellites is 900 km, as shown. from ka-12’s perspective, the angle between the ufo and sal-1 is 60°. from sal-1’s perspective, the angle between the ufo and ka-12 is 75°. image of triangle with ka-12, sal-1, ufo, angles 60° and 75°, side 900 km, and ? how far is the ufo from ka-12? do not round during your calculations. round your final answer to the nearest ten kilometers. blank km

Explanation:

Step1: Find the third angle of the triangle

In a triangle, the sum of interior angles is \(180^\circ\). Let the angle at UFO be \(C\), angle at KA - 12 be \(A = 60^\circ\), angle at SAL - 1 be \(B=75^\circ\). Then \(C=180^\circ-(60^\circ + 75^\circ)=45^\circ\)

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\), where \(a\), \(b\), \(c\) are the lengths of the sides opposite to angles \(A\), \(B\), \(C\) respectively. Let the distance between SAL - 1 and UFO be \(a\), distance between KA - 12 and UFO be \(b\), and distance between KA - 12 and SAL - 1 be \(c = 900\) km. We want to find \(b\) (distance from UFO to KA - 12), and the side \(c\) is opposite angle \(C = 45^\circ\), side \(b\) is opposite angle \(B=75^\circ\). So from \(\frac{b}{\sin B}=\frac{c}{\sin C}\), we can solve for \(b\):
\(b=\frac{c\times\sin B}{\sin C}\)
Substitute \(c = 900\), \(B = 75^\circ\), \(C=45^\circ\)
We know that \(\sin75^\circ=\sin(45^\circ + 30^\circ)=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\), \(\sin45^\circ=\frac{\sqrt{2}}{2}\)
So \(b=\frac{900\times\frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{2}}{2}}=\frac{900\times(\sqrt{6}+\sqrt{2})}{4}\times\frac{2}{\sqrt{2}}=\frac{900\times(\sqrt{6}+\sqrt{2})}{2\sqrt{2}}=\frac{900\times(\sqrt{3}+ 1)}{2}\) (rationalizing the denominator: \(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{2}}=\sqrt{3}+1\))
\(b = 450\times(\sqrt{3}+1)\approx450\times(1.732 + 1)=450\times2.732 = 1229.4\)
Rounding to the nearest ten kilometers, we look at the units digit \(9\), since \(9\geq5\), we round up the tens digit. So \(1229.4\approx1230\)

Answer:

\(1230\)