# find the radius of this circle

• Nov 13th 2005, 06:28 AM
Natasha
find the radius of this circle
A, B, C, D are points on a circle of centre O and TA is the tangent to the circle at A.

TA = 7.1 cm

Basically if we were to draw this circle we would first draw (Clockwise) the point D on the circle then A then C and then B.
Point T being outside of the circle slightly on the right of it.

Now angle BDA = 36 degrees and DAT = 28 degrees

From this info I need to find the radius of the circle and say if OT bisects AD?

I have demonstrated (using the circle property) that angle BOA = 72 degrees and BCA = 144 degrees and also OAD = 62 degrees.

But how can I find the radius and prove the bisection? Please any hints or answers to this treaky one :-)
• Nov 13th 2005, 07:17 AM
hemza
Usually I don't use the things already proven by the question poster (since I don't know if it is good and I don't want to give you wrong indications) but then I need to know information about C. (You do not give much info about C)

So what do you know about C ?

If you want me to use what you have already proven well just reply to this by saying so.

thank you
• Nov 13th 2005, 09:08 AM
Natasha
well C is on the circle, it's between B and A and forms an angle of 144 degrees. I am absolutely 200% confident about my answers so just use them as if they were part of the problem.
• Nov 13th 2005, 11:51 AM
hemza
ok,

thanks for you precalculated angles. I posted a figure with just the necessary information.

You almost did everything. The thing I will point is something very useful but that we forget to look at : Equal radiuses (plural of radius).

so OD=OA (each is a radius) so OAD is an isosceles triangle. So angles ODA=OAD=62 degrees so angle AOD=180-62-62=56 degrees. By sin law :

sin(56)/4.5 = sin(62)/OA so OA = radius = 4.79

OTA is a right-angled triangle. We have TA=7.1 and OA=4.79. I name H the intersection of TO with DA. We look for angle OHA equal or not to 90.

For that we begin by finding TOA. tan(TOA)= 7.1/4.79 so TOA = 56 (=HOA). So TOA=AOD ???!!!

So H is D.

Maybe you made a mistake. Maybe TA=4.5 and DA=7.1.
• Nov 13th 2005, 01:44 PM
Natasha
Thanks for that Hemza... Just a bit confused. First no I doubled check and I am sure about AT and AD lengths.

But look how I did it, and got a different answer. Tell me where I went wrong please or if you got it wrong?

Here it is

I started by dividing AD in two with a point E. Now from the triangle OAE we get OA=AE/cos(62) which is 4.79 cm so far so good but here its where it gets strange...

I drew the bisector of AD from O to the line AT, and called the point it makes on line AT point F. The projection of AF onto AD is AE, from this we get AF=AE/cos(28)=2.55 cm, which is not equal to AT and thus OT does not bisect AD (OF does)?

Who is right? Please explain why?
• Nov 15th 2005, 08:05 AM
hemza
our answers do not contradict each other because I also proved that it would not bisect

I proved that OT was too far (and at the left) from the potential bisector and you proved that the bisector was at the right of OT so it is the same thing.

The only difference is that the calculations I did permitted to see that OT is on OD so it is clear that it does not bisect