Week 18

Apply Pythagoras' theorem and trigonometric ratios to find angles and lengths in right-angled triangles {and, where possible, general triangles} in two {and three} dimensional figures. 

Here we have a quick look at some key trigonometric concepts.

MONDAY: It is easy to fall into the trap of thinking that trigonometric functions, in this case the sine function, are linear. If that were the case, then the suggestion in this task, that sin 70˚ = 0.966, would be true. In fact sin 70˚ = 0.940 (to 3 decimal places).

It is worth pointing out that for 'smallish' angles, the sine function is quite close to being linear. Consider this set of values. If one focusses on the first two decimal places, it is only when we get to 36˚ that it becomes clear that the values are not linear.

One way to counter Tricia's claim is to use an ad absurdum argument: If the pattern were linear, then sine 80˚ would equal 1.066 which is greater than 1.

A more grounded argument involves adding vertical lines like those below to the given diagram. If we think of the circle as having a radius of 1 unit, then the vertical lines' lengths give us the sine of the angles. As we can see, as the angles steadily increase (towards 90˚), the lines increase in ever smaller steps.

TUESDAY: This task exploits the idea that we can compare angles if we embed them in triangles that have something in common, for example if the triangles are both right-angled.

An informal, qualitative, way to solve this task is to imagine rotating the plane containing the red lines about the right-hand vertical edge of the cuboid (see the diagram below, left). As the horizontal red line overlaps a horizontal edge of the cuboid, the slanting red line intersects a horizontal edge of the top face between its two endpoints rather than at an endpoint. So the red slanting line is steeper than the green slanting line which means the red angle is larger than the green angle.

A more analytic way is to form the right-angled triangles shown below, right. The triangles have the same height, but the red triangle has a smaller base (the red and green bases are in the ratio 1 : √2). So the red angle is larger than the green angle.

You might want to make the trigonometric ideas more explicit by using this variant:

 

TUESDAY: Back to basics....

A grounded way of solving this is to consider the right-angled triangle where the length of the hypotenuse is 1 unit and the side opposite our angle is 0.6600 units. Then the length of the adjacent side, a units, is given by a² = 1 – 0.6600², so a = 0.7513 and so the tangent is 0.6600÷0.7513 = 0.879 (correct to 3 dp).

An alternative would be to use the relations sin²A + cos²A = 1 and tanA = sinA/cosA, though students are unlikely to know these. It is also heavy handed.

THURSDAY: This is quite a complex task, though its underlying aim is relatively simple. This is to convey the idea that, at this level of schooling, trigonometry is essentially a form of scale drawing (albeit one that is restricted to right-angled triangles) where someone has already done the drawing and recorded the measurements for us.

The task involves several steps. One way to break it down is like this: enlarge the original figure by a scale factor 0.2 to give the figure below-left. Then rotate, enlarge and re-attach the smaller (brown) triangle, as in the figure below-right, so that its side that had become 20 units long becomes 25.66 units long - an enlargement of scale factor 1.283. So the side that had become 23.84 units long becomes 30.59 units long.

FRIDAY: Another trigonometry-as-scale-drawing task....

This again involves several steps. One way to break it down is shown in the figures below. These involve an enlargement of scale factor 0.6 and then an enlargement (of just the rotated brown triangle) of scale factor 76.98 ÷ 71.52 = 1.076.