# Wind Turbine Problem

The ‘wind turbine problem’ is a gem from the 2014 Ordinary Level Irish Leaving Certificate exam, and one I love working over with students. In terms of points it is pretty juicy (a 50 pointer), but there is nothing in it too difficult. It is made up of 3 sections, with a total of 6 individual questions.

8.a.i Is nothing more than observation. The wind turbine is made up of three blades, each originating at a central hub. When the turbine turns the blades move in a circular motion. So we are thinking about a circle with three lines of radii.

We are told that the blades are “equally spaced,” meaning that the angle between each line is exactly a third of the number of degrees in a full circle:

360 / 3 = 120
120°

We’re asked to find the area of the circle drawn by one rotation of the blades. This is asking us to imagine a circle with a radius of 65 metres (remembering that 65m is the size of the blades).

The equation of the area of the circle is Area=πr2. So let’s put that to some use:

Area=πr2
π(652) = Area
4225π ∨ 13,273.22869
13,273 m2 – to the nearest whole number

Here we are asked to construct a triangle by treating the three points where the blades meet the line of the circle as the points of the triangle (shown here in red):

Finding the area of this triangle is not difficult, but it is complex. So we have to pay close attention to what we’re doing. Okay, we see that the whole triangle is made up of three smaller triangles, all sharing a common point at the centre of the circle. As every line coming from the centre is a radius to the same circle, they are all 65m in length. This means that each of these smaller triangles is isosceles and the large is equilateral.

Let’s take a closer look at one of these triangles, remembering that if we find its area we must multiply that by 3 – as the larger triangle is made up of all three smaller triangles.

At this point it would be useful to draw a line from the corner with the known angle, perpendicularly through its opposite line. This gives us a right angle to work with and makes things much easier. But we must remember that we are now dealing with 16 of the original triangle.

We know that to find the area of a triangle we use the equation 12b×h. On this triangle the base will be the line opposite the 60° angle, and the height its adjacent. This allows us to use some trigonometry; the equations Sin60 = opp65 and Cos60 = adj65. Let’s do it:

Sin60 = opp65
65Sin60 = opp ⇒ 65√(3)2

Thus:

12b×h
(12(65√(3)2)) × 652
914.7393327

We remember that this triangle is 16 of the larger triangle, therefore:

6(914.7393327)
5488.435996 ∨ 5488m2 to the nearest whole number.

15 times per minutes
900 times per hour
21,6000 times per day (24 hours)
7,884,000 times per year (ignoring leap years)
197,100,000 per year

0.31(197,100,000)
61,101,000

In the form a×10n, where 1 ≤ a < 10 and n ∈ Z (ie. Scientific Notation):

6.1101×107

All we are looking at here is another simple piece of trigonometry. We see this more clearly when we draw out what has been described:

In the case of this right angle we know both the value of the adjacent line (100m) and the angle of θ (60°). So we can use the equation tanθ = oppadj:

tan60 = opp/100
100tan60 = 173m

This is just a matter of going back and reworking the trigonometry: