Question

jan and ronnie are standing 10 miles apart. both can see a kite flying in the sky. the angle of elevation from jan to the kite is 35 degrees and the angle of elevation from ronnie and the kite is 45 degrees. how far is the kite from jan?

Explanation:

Step1: Find the third angle of the triangle

The sum of angles in a triangle is \(180^{\circ}\). Let the angle at the kite be \(C\). Then \(C = 180-(35 + 45)=100^{\circ}\)

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Let \(a\) be the distance from Ronnie to the kite, \(b\) be the distance from Jan to the kite (\(x\)), and \(c = 10\) miles (distance between Jan and Ronnie). So \(\frac{x}{\sin45}=\frac{10}{\sin100}\)

Step3: Solve for \(x\)

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Answer:

The kite is approximately \(7.19\) miles from Jan.