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an officer spots two boats with an angle of 53° between them. boat a is…

Question

an officer spots two boats with an angle of 53° between them. boat a is 2.3 km away, and the angle at boat a between the officer and boat b is 76°. how far is the officer from boat b? distance from boat b = ? km round your answer to the nearest hundredth.

Explanation:

Step1: Find the third angle of the triangle

In a triangle, the sum of angles is \(180^\circ\). Let the officer's position be \(O\), Boat A be \(A\), and Boat B be \(B\). We know \(\angle AOB = 53^\circ\) and \(\angle OAB = 76^\circ\). So \(\angle OBA=180^\circ - 53^\circ - 76^\circ = 51^\circ\).

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Here, we want to find the distance from the officer (\(O\)) to Boat B (\(OB\)), let \(OB = x\), \(OA = 2.3\) km, \(\angle OAB = 76^\circ\), \(\angle OBA = 51^\circ\). So \(\frac{x}{\sin 76^\circ}=\frac{2.3}{\sin 51^\circ}\).

Step3: Solve for \(x\)

\(x=\frac{2.3\times\sin 76^\circ}{\sin 51^\circ}\). Calculate \(\sin 76^\circ\approx0.9703\) and \(\sin 51^\circ\approx0.7771\). Then \(x=\frac{2.3\times0.9703}{0.7771}\approx\frac{2.23169}{0.7771}\approx2.87\) km.

Answer:

\(2.87\)