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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

a triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6cm respectively. find the sides AB and AC.

friendship is tan 90°.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

If your problem is like the drawing below.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

yes it is the right drawing and the answer is correct too.

I am not able to solve it.

pls. elaborate it with the process.

friendship is tan 90°.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

we need to find the sides by using simple theorems like the lengths of two tangents from an external point to a circle are equal.

I used that but was not able to get the full length of AB and AC..

I am having maths exam tomorrow, pls help soon.

friendship is tan 90°.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,084

hi niharika_kumar

Hope this is in time to be helpful. My method is convoluted but it will help you revise lots of techniques so that may be useful. Diagram below.

AG, BG and CG are the angle bisectors and GD is perpendicular to BC.

As you know BD and GD you can work out the sin and cos of angle GBD.

then use sin2x = 2sinxcosx to get the sine of angle B

Similarly you can get sin C.

As the angles of the triangle add to 180

sin A = sin (B+C) = sinBcosC + sinCcosB .

So you can use the sine rule to get AB and AC.

I would leave the square roots in the expressions in the hope they'll cancel out at some stage.

If I think of a quicker method, I'll post it.

Good luck with the exams.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

thnx for the wishes:)

but i am not able to really get the method:(

friendship is tan 90°.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,084

hi Niharika

How did the exams go? Are they all finished for now?

I did try to find a simpler method for this but without success. If you look at

http://www.mathisfunforum.com/viewtopic.php?id=20409

you'll see that bobbym had a go at this using geogebra, and then with a formula.

I then posted a more detailed account of my method.

It uses some trig. angle formulas which you may not have met yet, which is why I tried to find another way. The formulas I've used are shown at

http://www.mathsisfun.com/algebra/trigo … ities.html

http://www.mathsisfun.com/algebra/trig-sine-law.html

If you haven't met those yet, then I'm mystified as to how your teacher expected you to do this question.

Even if you have met them, it's still quite a long trail to get the result.

My starting point is to use the right angled triangle shown to get the angle GBD.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

now I understand your method.

thanks for that

yes my exams are over now and it was good.(This question luckily didn't come )

I could solve all the questions in maths (just hoping that I have not committed any silly calculation mistake).

friendship is tan 90°.

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

my maths teacher solved it by dividing the figure into several triangles and using heron's formula.

friendship is tan 90°.

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My maths teacher does that the same way. Besides, that is the method prescribed in Monica Capoor's book

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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