Week 19

Know the exact values of sin 𝛉 and cos 𝛉 for 𝛉 = 30˚, 45˚, 60˚ and 90˚; know the exact value of tan 𝛉 for 𝛉 = 0˚, 45˚, 60˚.
{Know and apply the sine rule and the cosine rule to find unknown lengths and angles.}
{Know and apply Area = ½ab sinC to calculate the area, sides or angles of any triangle.} 

Here we have collected together three statements of 'knowledge, skills and understanding' and are addressing them in just one week's set of tasks. This is not ideal, but we're running out of weeks!

So we have a very quick look at some familiar trigonometric ratios and at the sine, cosine and Area-of-a-triangle rules. The tasks can generally be solved without formal trigonometry, by for example using Pythagoras' Rule and ideas about similarity. An aim of the tasks is to remind students of the origins of the formal rules, which can be done in various ways. For example, you might explicitly ask students to solve a given task in two ways - by using and by not using one of the rules. Or you might leave it open, to see what methods students spontaneously adopt, which can then be shared.

MONDAY: This task doesn't require the explicit use of trigonometric ratios, but it rests on the fact that in a right-angled triangle, if the other angles are 45˚, the triangle is isosceles, and if the other angles are 30˚ and 60˚, the length of the hypotenuse is twice the length of the shortest side. In turn, this means that the sides are in the ratio 1 : 1 : √2 and 1 : √3 : 2, using Pythagoras' Rule.

We can find the lengths of sides in the first two triangles in a step by step way (below, left) or in a more curtailed way by making direct use of the above ratios (below, right).

We could now keep repeating either of these processes. Or we could curtail things by making use of the fact that the hypotenuse of length 20√6 mm that we have just found is ½√6 times the length of the first triangle's hypotenuse of 40 mm. So if we apply this multiplier 3 more times, we get to the final hypotenuse and the value of x:
x = 20√6 × ½√6 × ½√6 × ½√6 = 20 × 36 ÷ 8 = 90.

TUESDAY: The sine rule applies to triangles. It has a very simple proof based on writing equivalent expressions for a triangle's height. We can do the same kind of thing for this special quadrilateral.

 WEDNESDAY: We can express all the key lengths in the given figure using just two variables (once we have used our knowledge of 60˚ right-angled triangles). The resulting expressions for the yellow and green areas can then be equated and simplified using Pythagoras' Rule, until the desired result falls out.

The result also falls out very neatly using the cosine rule. Consider the diagram below, where we have added two red diagonal lines that serve as hypotenuses. The total green area is equal to a² (Pythagoras' Theorem). Using the cosine rule, we can write
a² = b² + c² – 2bc cos120 = b² + c² + 2bc×½ = b² + c²  +  bc
= area of yellow squares + yellow rectangle.

As alluded to earlier, we can use first principles and write the area of the green squares as
(b + ½c)² + (½c√3)² = b² + bc + ¼c² + ¾c² = b² + c² + bc = yellow area.

WEDNESDAY Extra: Here is a variant of the above task where the the hypotenuses are already present. The task might be useful for students who are stuck on the previous task or to consolidate the earlier ideas.

We can express the area of the yellow square using Pythagoras' Theorem and the green square using the cosine rule. Or we can use first principles to find expressions for the sides of the squares (namely 6√2 cm and 6√3 cm). It turns out that the difference between the areas of the green and yellow squares is 36 cm². Put another way, the area of the green square is equal to the area of the two yellow squares, below:

 

THURSDAY: In this task it is fairly easy to see how the values of a and b change as C moves along the green arc. However, it is less easy to be certain about their product.

As point C approaches B the product ab approaches 0 since b approaches the fixed value c and a approaches 0. But what happens when C is say one-third and one-half of the way along the green arc?

One way to achieve certainty is to apply the area rule to triangle ABC:
Area of triangle ABC = ½ab sinC.

The angle C doesn't change when C moves on the green arc, so the maximum value of ab is achieved when the triangle's area is a maximum. If we take c as the base (which also doesn't change), this occurs when C is exactly halfway between A and B since the height of the triangle is then a maximum.

 

FRIDAY: Here we have two triangles with the same size base. Their third vertices are the endpoints of  radii of the same circle. How do the angles that these radii make with the base affect the triangles' heights?

The heights of the two triangles can be expressed as OPsin25˚ and OQsin50˚, or Rsin25˚ and Rsin50˚, where OP = OQ = R. The sine function is not linear, as we saw in Task 18A: it turns out that the value of sin50˚ is slightly less than 2sin25˚, and so the area of the larger triangle is slightly less than twice that of the smaller triangle.

We can say all this in condensed form using the area formula:
Area of AOP = ½rRsin25˚, area of BOP =½rRsin50˚ (where r is the radius of the smaller circle);
 2×½rRsin25˚ > ½rRsin50˚ because 2sin25˚ > sin50˚.