As observed from the top of a lighthouse, 100 metres above sea level, the angles of depression of two ships are 30 degrees and 45 degrees. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. Use the square root of three is equal to 1.732.
First, let’s sketch this out. We have a lighthouse. We’ll put our observed distance here at the yellow dot. That observation distance is 100 metres above sea level. The lighthouse makes a right angle with sea level. And we have two ships: one behind the other. From our observation point, we need to create two right triangles, here and here. But how do we label the angles of depression?
If we draw a line from the observation point that runs parallel to sea level, the first angle of depression is 30 degrees. This diagonal line intersects the two parallel lines of sea level and the line of observation. And the created angle by the boat further out is 30 degrees. Here’s the first created right triangle.
For our second right triangle, we follow the same procedure. This time, we have a 45-degree angle of depression. We can label the interior angle by the closer boat 45 degrees as well. These are alternate interior angles. And here we’ve created our second right triangle.
Before we move on, let’s be really clear about what distance we’re solving for. We are interested in the distance between the two ships. In order to find that, we’ll take the distance of the boat that’s furthest away. And then, we’ll subtract the distance of the boat that’s closest to the lighthouse.
In order to do that, we’ll need to calculate these two missing side lengths. First, let’s focus on the pink triangle — on the boat that’s furthest away. We know an angle, an opposite side length, and we’re interested in the measure of the adjacent side length.
In a right triangle, dealing with opposite and adjacent side lengths is the tangent ratio. So we’ll set up a tangent ratio for the pink triangle. Tangent of 30 degrees equals the opposite side length 100 over the missing adjacent side length. We’ll call it 𝑥.
We need to solve for 𝑥. To do that, we multiply by 𝑥 over one on the right and the left. The left side now says 𝑥 times tangent of 30 degrees on the right the 𝑥s have canceled out, leaving us with 100. From here, I divide the left side by tangent of 30 degrees. And if we do that on the left, we need to also divide by tangent of 30 degrees on the right.
We then see 𝑥 is equal to 100 over tangent of 30 degrees. We know that the tangent of 30 degrees equals the square root of three over three. We can substitute the square root of three over three for tangent of 30 degrees. The new statement says 𝑥 equals 100 divided by the square root of three over three.
We know that dividing by a fraction equals multiplying by the reciprocal. 100 times three equals 300 over the square root of three. Well, what we want to do now is get rid of the square root of three in the denominator. We do that by multiplying the numerator and the denominator by the square root of three.
The numerator we’ll say 300 times the square root of three. And in the denominator, the square root of three times the square root of three equals three. 300 divided by three equals 100. So our 𝑥-value is equal to 100 times the square root of three. For now, we’ll keep 𝑥 as 100 times the square root of three.
Moving on to the small triangle, fortunately, for us, we notice something about this triangle. This is a 45, 45, 90 triangle. And that means the side lengths across from the 45 degrees are the same lengths.
Our second boat is 100 metres from the lighthouse. And the furthest boat away is 100 times the square root of three metres away. To find the space between them, we need to take the large distance and subtract the small distance: 100 times the square root of three minus 100. If we take out the factor 100, we now have 100 times the square root of three minus one.
We will now use our approximation for the square root of three. 100 times 1.732 minus one is equal to 100 times 0.732. Subtracting the two distances gives us 73.2 metres.