tangents to circles

these are based on problems from the CBSE exams (India) Y10

circle theorems meet 0.5absinC

an idea by Martin Wilson, in Harrogate, blending the circle theorems with area

a ppt is here

you could use normal trigonometry (by bisecting chords) but Martin's intention is that students use the (more efficient) method of calculating 0.5 a b sinC

resources 2, 3 and 4 present an interesting relationship between sines
that is not easy to justify(?)
without involving the trig. formula sinA + sinB = 2sin(semi-sum).cos(semi-difference)
and noting that cosD = sin(90 - D)

there's a sequence to follow

circle theorems using parallels

normally the circle theorems are justified using isosceles triangles
but it's quite neat to use alternate and corresponding angles on parallel lines

this and other ideas are explored more substantially by Dietmar Kuchemann in this article

construct a parallel line

but A = B (isosceles triangle) so the angle at the entre is double the angle at the circumference


the angle in a semicircle being 90 degrees follows from the fact that an isosceles triangle is perpendicularly bisected by its line of symmetry

construct a parallel line














 

measuring angles on circular grids

Paul Andrews suggests, in an article reprinted by nrich, that measuring angle tasks can be fairly dull, even when students have been asked to estimate their sizes first

Paul suggests drawing triangles connecting dots on various point-circles

5, 6 and 9 dot circles produce angles that are all integer

students can be asked to draw all the different triangles they can find on e.g. 9 dot circles

and then measure the three angles

they need to draw these as accurately as they can
















although it isn't intended to be part of this task, some students might note that equal angles subtend equal arcs

there are:
2 different triangles for a 5 point circle with angles that are M(36)
3 different triangles for a 6 point circle with angles that are M(30)
7 different triangles for a 9 point circle with angles that are M(20)

the circles are drawn fairly large to aid accurate measurement with a protractor:

perps

draw two lines, AB and AC
with any point, D, 'between' them

drop perpendiculars from D to AB (at F) and from D to AC (at E)

join E to F (the feet of these perpendiculars)

then create a perpendicular from A to EF (meeting EF at G)


possibly set this up (e.g. with GeoGebra)

establish, by measuring and dragging D around
that angle GAE = angle FAD

try to prove it

can students find another angle the same as these?

cyclic quadrilateral angles

show that if the angles in a quadrilateral are in a linear growing sequence (i.e. in arithmetical progression) then it must be cyclic

while you're at it, show that such angles must either all be even or all odd

angle in a semicircle

there's a neat proof of this circle theorem in Paul Lockart's article
('A Mathematician's Lament')









Paul's article is worth a read, or a re-read, arguing that maths is fundamentally about problems and these must be made the focus of a student's mathematical life, however painful and frustrating this may be

he describes a student's argument that if you take a right-angled triangle (hypotenuse passing through the centre) and rotate it around (through 180) it creates a four-sided shape, the sides of which must be parallel ...
so it is a parallelogram

further, the diagonals are both the same length ...
so it must be a rectangle

they complete their view that since the triangle got rotated halfway round the circle, one 'tip' ends up exactly opposite from where it started
that's why the other diagonal is also a diameter

and since it's a rectangle the angles are 90 degrees

going off at a tangent

why is the tangent to a circle at right angles to the radius?
two methods for establishing this both involve an idea of limit

one way is to establish (RHS) that if you join the middle of a chord to the centre of the circle the line is perpendicular to the chord and then move this chord outwards (parallel to itself) until it just about leaves the circle...

another involves using a chord, extended beyond the circumference at both sides

you can easily show that the two angles that the chord makes with the radiuses are equal (RHS again)
so the supplements (other angle on the straight line) to these angles are equal

again, moving the chord steadily out of the circle shows that these two angles become 90 degrees when the chord becomes a tangent