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).]]>

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

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

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

]]>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.]]>

]]>

I am not able to solve it.

pls. elaborate it with the process.]]>

If your problem is like the drawing below.

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