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### Dhaka Secondary 2009/8

Posted: Tue Feb 01, 2011 11:49 pm
For triangle $ABC$ $\angle C$ is $90^{\circ}$. $\angle BAC$ is $30^{\circ}$ & $AB$ is $1$cm. $D$ is a point within $ABC$ so that angle $\angle BDC$ is $90^{\circ}$ & $\angle ACD = \angle DBA$. $AB$ & $CD$ meets at $E$. Find $AE$.

### Re: Dhaka Secondary 2009/8

Posted: Tue Feb 01, 2011 11:55 pm
ans is $\frac{1}{2}$. i just used similiar triangle properties and general trigonometric means.

### Re: Dhaka Secondary 2009/8

Posted: Wed Feb 02, 2011 12:02 am
^By outline I did not mean that you should just post what you used to solved the problems (it is more or less apparent that some Euclidean geometry and trig can solve almost any problem at this level). You better post what you used to solve what.

It is also best for everyone if you post complete solution. Remember, in National we need to write complete solutions.

### Re: Dhaka Secondary 2009/8

Posted: Wed Feb 02, 2011 12:04 am
tahmid, full sollution or at least a little hint.. Please

### Re: Dhaka Secondary 2009/8

Posted: Wed Feb 02, 2011 12:05 am
sorry i'll try from next time and post almost full but not formal solns

### Re: Dhaka Secondary 2009/8

Posted: Thu Mar 22, 2012 8:20 pm
Plz give the full solution or a hint

### Re: Dhaka Secondary 2009/8

Posted: Fri Jan 31, 2014 9:11 pm
in triangle ABC, cos 30=AC/1
ANGLE DCB+ ANGLE DBC=90
ANGLE ACD+ANGLE DCB=90
GIVEN, ANGLE ACD=ANGLE DBA
SO, ANGLE ACD=ANGLE DBA
ANGLE DBC=ANGLE DBA=60/2=30
sin(120)/AC=sin(30)/AE
AE=0.5

### Re: Dhaka Secondary 2009/8

Posted: Fri Jul 22, 2016 12:47 pm
We can easily show that triangle FCD and triangle EDB are similar . So FD/DE = CD/BD = FC/BE . Again we can see that triangle FCD and triangle BCD are similar . Again FD/CD = CD/BD = FC/BC .If we merge the conclusions we get FD/CD = FC/BC = FD/DE = CD/BD = FC/BE. We have FC/BC = FC/BE hence BC=BE. Now we can apply trigonometry and get the value of BE then i think we all know what to do

[Q.E.D]

### Re: Dhaka Secondary 2009/8

Posted: Mon Sep 05, 2016 3:55 pm