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G.c.s.e Circle With Tangent Question

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kuiperbelt | 13:50 Thu 30th Mar 2023 | Science
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No need for trig, just similar triangles. Diagram described by zebo: https://ibb.co/z7B8p06 Right-angled triangle ONA is similar to right-angled triangle OAB (both right-angled and angle AON common to both). So OB/OA = OA/ON Rearranging gives OB = OA²/ON = 41/4 = 10.25
15:31 Thu 30th Mar 2023
foo I cdnt see the q so I was gonna say
foo it seems we are - - - going around in circles
ter daah

I will look now
Easiest route seems to be to deduce the radius, and the y-coordinate of A. Then the value of the angle AOB can be found by taking the tangent of the coordinates of point A. Then the value of length b (from O to B) follows from cos (AOB) = r/b.

The rest is detail, somebody else can do the numberwork.
Radius is just straight root of 41?
Square root of 41
radius is 41
OAB is 90
A perpendicular line from A onto OB at N gives another right angled triangle, hypotenuse OAis sq rt of 41, from the equation, side ON is 4

The rest of it is trig, using cos to find angle AON then using that in the triangle OAB, OA/OB = cos AON.
Sorry, bobbin; I missed out the words "square" and "root" and " of".

.
I get 10.25.

It became much easier once I read the question properly and saw the equation at the top! As I used to tell my classes, When all else fails read the bloody question properly.
Yes 82 divided by 8
The coordinates of B, to 1 decimal place, are (10.3, 0.0)
No need for trig, just similar triangles. Diagram described by zebo:
https://ibb.co/z7B8p06

Right-angled triangle ONA is similar to right-angled triangle OAB (both right-angled and angle AON common to both). So
OB/OA = OA/ON
Rearranging gives
OB = OA²/ON = 41/4 = 10.25
That's a much lovier solution than my "just throw trig at it" approach.
Does exam question contain a typo... should it say 1 decimal place not one decimal point.
Thats my contribution !
another way is to see that A must be (4,-5)
Then use pythogora's
The various solutions proffered are by no means exhausted.

Etch's 'Similar Triangles' are quite prepossessing and steer away
from the humdrum of trigonometry. This is not to suggest they exceed the pulchritude as found in for example, the Mandelbrot Set or Fractal Geometry. However, there is something both aesthetic and simplistic when a mathematical problem is borne out by illustration.

Perhaps the only thing preventing Etch from receiving the award of BA, 'Similar Triangles' might lie outside the remit of GCSE?

kuiper, I will demonstrate an alternative method if the student feels it is to their benefit.
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My grandson has pointed out that one answer does not require the use of an electronic calculator (EC) or tables. Even bobbinwales Pythagoras solution would eventually need tables or EC to look up square roots!

Therefore BA goes to Etch!! BTW Zebu, your alternate solution presumably would need EC or tables too??
// BTW Zebu, your alternate solution presumably would need EC or tables too?? //

No, it does not.
someone isnt a happy bunny! : )

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