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Posts Tagged ‘Triangle’

A nice little triangle puzzle

December 5, 2016 8 comments

Here is a nice little puzzle I saw from brilliant.org on Facebook.

Have you worked it out yet?

Here’s what I did:

First I drew a diagram (obviously).

And worked out the area of the triangle.

Then the area of each sector.


Leaving a subtraction to finish.

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Categories: #MTBoS, Maths, Starters Tags: , ,

Exact Trigonometric Ratios

November 8, 2015 4 comments

This morning I read this interesting little post from Andy Lyons (@mrlyonsmaths) which looked at teaching the exact Trigonometric Ratios for certain given angles (namely 0, 30, 60, 90 and 180 degrees). The post gave a nice little info graphic linked to the unit circle to show what was going on and then focused on methods yo remember the ratios.

While reading it I thought about how I introduce these exact Trigonometric Ratios. I first like to know that my students have a thorough and in depth understanding of right angled triangles and the trigonometry involved with them (including Pythagoras’s Theorem). I feel this is imperative to learning mathematics, the Triangle is an extremely important shape in mathematics and to fully understand triangles you must first fully understand the right angled triangle. The rest follows from that.

Once these are understood then you can move on to the trigonometric graphs, showing how these can be generated from right angled triangles within the unit circle, as shown in the info graphic on Andy’s post. Once the graphs are understood then the coordinates f the x and y intercepts and the turning points give us nice exact values for angles of 0, 90 and 180 degrees. This leaves us with 30, 60 and 45 to worry about.

At this point I introduced 2 special right angled triangles. First up is the right angled isosceles triangle with unit lengths of the short sides. This obviously gives us a right angled triangle that has two 45 degree angles (as the angle sum of a triangle is 180) and a hypotenuse of rt2 (via Pythagoras’s Theorem).

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Using our definitions of trigonometric ratios (ie sin x =opp/hyp, cos x = adj/hyp and tan x = opp/adj) we can clearly see that tan 45 = 1 and that sin 45 = cos 45 = 1/rt2. This aids the understanding more than just giving the values and allows students a method of working these values out easily if stuck.

The second triangle is an equilateral triangle of side length 2 cut in half. This gives us a right angled triangle with hypotenuse 2, short side lengths 1 and rt3 (again obtained through Pythagoras’s Theorem) and angles 30, 60 and 90.

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Again we can use our definitions of trigonometric ratios to conclude that sin 30 = cos 60 = 1/2, sin 60 = cos 30 = rt3/2, tan 30 = 1/rt3 and tan 60 = rt3.

This is again good for deeper understanding and for seeing why sin x = cos 90 – x, and cos x = sin 90 -x. This can lead to a nice discussion around complementary angles and that the word cosine means “sine of the complementary angle”. This triangle is also a good demonstration that tan x = cot 90 – x, when you come to higher level trig.

UKMT Maths Challenge Q16

November 7, 2014 1 comment

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Above is a photo of question 16 from yesterday’s UKMT Senior Maths Challenge. And what I fantastic question it is!

It shows (incase you can’t make it out on the picture) a rectangle and a circle which have the same centre. The two shorter sides of the rectangle are tangents to the circle, and the rectangle is 6×12. The task is to find the area that is inside both.

After the challenge had been completed one if my Y13s came to ask about it, and it took me a little while to work it out. I played around with some similar triangles, thinking that I would need to calculate the bit of the rectangle outside the circle and subtract them from 72. Then I had the realisation that I could easily split it up into triangles and sectors that I could calculate the area of with great ease.

First I sketched:

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Showing the rectangle split into areas and sectors. I then considered the triangles, if I split then into right angled triangles I had a hypotenuse of 6 and a height of 3, so the cosine of the top angle must be 0.5, which means the angle must be pi/3.

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I then used the sine of this angle ((rt3)/2) to calculate the opposite side as 3rt3. Then thought “why didn’t I use Pythagoras’s Theorem?!” A quick check using Pythagoras’s Theorem shows us the side is rt27, which simplifies to 3rt3. The area of the triangle then must be 9rt3.

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Knowing that the angles at the centre in the two triangles are 2pi/3, that angles round a point add up to 2pi and that the sectors are congruent means we can deduce the angle if each sector is pi/3. We also know the area of a sector is (r^2(theta))/2, which gives the area of each sector as 6pi.

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Finally we sum the areas and get 18rt3 + 12pi.

The student had looked at the answers and estimated them using the knowledge that rt3 is around 1.7 and pick is a but more than three. This meant he could eliminate c,d and e, but he didn’t get to the final answer.

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