A boat, stationed at the North of a lighthouse, is making an angle of

This question involves three dimensions (not at the same time), and the north and east directions are not useful until the last step.

First, let us take the sea level view. From the first sentence, we can draw a 30-60-90 triangle with the shorter leg being 300 feet. Then we can figure out the distance of the boat, it is \(300*\sqrt{3} = 300\sqrt{3}\) feet.

Next, we use the lighthouse height to calculate how far the east boat is away from the lighthouse. The east boat makes a 45-45-90 triangle so the distance is simply \(300\) feet as well.

Finally, let us go to the bird's eye view. The lighthouses are \(300\sqrt{3}\) feet and 300 feet away, and they make a right angle. Using the Pythagorean theorem they have a distance of \(\sqrt{300^2*3 + 300^2} = 300*\sqrt{3 + 1} = 600\) feet.

Ans: D
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Step 1: Understanding the question
Using the relation of special right angle triangle 30° : 60° : 90° as x : x√3 : 2x
The boat stationed at the North of a 300-feet high lighthouse, making an angle of 30° with the top of the lighthouse, is 300√3 feet away from the lighthouse (opposite to 60°).

Using the relation of special right angle triangle 45° : 45° : 90° as x : x : x√2
The boat stationed at the East of the same lighthouse, is making an angle of 45° with the top of the lighthouse, is 300 feet away from the lighthouse (opposite to 45°).

Step 2: Calculation
The two boats are making right angle with each other, hence using Pythagorean theorem to find distance between them.
Distance between the two boats = √[\((300√3)^2 + (300)^2\)] = 300 * √[\((√3)^2 + 1^2\)] = 300*2 = 600 feet

D is correct
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Hello! For some reason I am not understanding "\sqrt{300^2*3 + 300^2} = 300*\sqrt{3 + 1} = 600[/m] feet" part. Wouldn't the square and the square root cancel each other out which makes it 300*3+300=1200? I know this is wrong but can't figure out why. Thank you in advance!

Solution:

Consider the 2 figures:



\(AB\) is the right house \(300\) feet long. A boat is located at \(C\) making \(30°\) with the top of the lighthouse and a boat is at \(D\) making an angle of \(45°\) with the top of the lighthouse.

We need to find the value of \(DC\).

\(\triangle ABC\) is a right angle \(30-60-90\) triangle whose side ratio will is \(1:\sqrt3 : 2\). So, if the side \(AB=300\), then the side \(BC=300\sqrt3\).

Similarly, \(\triangle BAD\) is a right angle \(45-45-90\) triangle whose side \(BD=AB\). So, if the side \(AB=300\), then the side \(BD=300\) aswell.

Now that we know the value of \(BC\) and \(BD\), we can get the value of \(DC\) using Pythagoras theorem.

\(DC=\sqrt{(300\sqrt3)^2+300^2}=600\)
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Hi lexjp1019 .

We have a sum under the square root \(\sqrt{300^2*3 + 300^2}\). We cannot simplify \(\sqrt{A + B}\) unless we calculate A + B. Here, I factored out \(300^2\) from both terms to get \(\sqrt{300^2*(3 + 1)} = \sqrt{300^2}*\sqrt{3 + 1}\). So you'll see \(\sqrt{3 + 1}\), and a 300 outside, which was a result of \(\sqrt{300^2}\)
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Thank you for your kind response!