Search found 1175 matches
- Mon Mar 28, 2011 5:55 pm
- Forum: Secondary Level
- Topic: sum of two sides multiplied on 2 opposite points
- Replies: 3
- Views: 3401
Re: sum of two sides multiplied on 2 opposite points
Right you are.........I didn't got much time to explain it plainly.........
- Sat Mar 26, 2011 2:31 pm
- Forum: Secondary Level
- Topic: sum of two sides multiplied on 2 opposite points
- Replies: 3
- Views: 3401
Re: sum of two sides multiplied on 2 opposite points
By applying sine law on $\bigtriangleup PSR$, $ \frac {PR} {sin \angle PSR} =2R$ Now the area of quadrangle $PQRS$ is $(PQRS)=(PSR)+(PQR)$ $=\frac {1} {2} .PS.SR.sin \angle PSR+ \frac {1} {2} PQ.QR. sin \angle PQR$ $=\frac {1} {2} sin \angle PSR(PS.SR+PQ.QR)$ $=\frac {1} {2} . \frac {PR} {2R} (PS.SR...
- Sat Mar 26, 2011 1:52 pm
- Forum: Geometry
- Topic: Proving Ptolemy's inequality with vectors
- Replies: 4
- Views: 4138
Re: Proving Ptolemy's inequality with vectors
But I think I will be posting the 'complex' one within a little time..........
- Sat Mar 26, 2011 1:49 pm
- Forum: Geometry
- Topic: transformers(cycli quadrilleterals in disguise)
- Replies: 1
- Views: 2121
Re: transformers(cycli quadrilleterals in disguise)
Are you sure your question is right? $\angle BAC + \angle BDC=\pi $ makes no sense, it implies $\angle BAC +\angle BCD=\pi$ which is not at all useful.I think the question should contain $\angle BAD + \angle BDC=\pi $ or$\angle BAC = \angle BDC $ to make quadrangle $ABCD$ a cyclic one.
- Sat Mar 26, 2011 1:19 pm
- Forum: Geometry
- Topic: Proving Ptolemy's inequality with vectors
- Replies: 4
- Views: 4138
Re: Proving Ptolemy's inequality with vectors
Yeah I know that of course.......but I thought in many cases,vector and complex is the same thing we call by two names.........as in vectors $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ is all the same as $(\alpha - \beta) +(\beta -\gamma )=( \alpha - \gamma )$ ,isn't it?
- Sun Mar 20, 2011 9:57 am
- Forum: Geometry
- Topic: Proving Ptolemy's inequality with vectors
- Replies: 4
- Views: 4138
Re: Proving Ptolemy's inequalit with vectors
Solution: $\overrightarrow{FD}\times\overrightarrow{CE}=(\overrightarrow{FC}+\overrightarrow{CD})\times(\overrightarrow{CD}+\overrightarrow{DE})$ $=(\overrightarrow{FC}\times\overrightarrow{CD}+\overrightarrow{FC}\times\overrightarrow{DE}+\overrightarrow{CD}\times\overrightarrow{CD}+\overrightarrow{...
- Sun Mar 20, 2011 9:42 am
- Forum: Geometry
- Topic: Proving Ptolemy's inequality with vectors
- Replies: 4
- Views: 4138
Proving Ptolemy's inequality with vectors
Try proving Ptolemy's inequality by vectors.
(I think this is useful for the inequality part $CD.EF+DE.CF\geq CE.DF$)
(I think this is useful for the inequality part $CD.EF+DE.CF\geq CE.DF$)
- Mon Feb 07, 2011 10:58 am
- Forum: Secondary: Solved
- Topic: Rangpur Secondary 2011/4
- Replies: 4
- Views: 4403
Re: Rangpur Secondary 2011/4
Your ans. is not complete,as the lower limit of $X$ is $4$,and the question is about finding the higher and lower limit of $X$photon wrote:can't it be 20>x>8....according 2 ABC??
- Fri Feb 04, 2011 6:34 pm
- Forum: H. Secondary: Solved
- Topic: Rangpur Higher Secondary 2011/2
- Replies: 5
- Views: 9489
Re: Rangpur Higher Secondary 2011/2
$(9^x)(9^{18})=16^x$
$\Rightarrow (9^x)((9^9)^2)=16^x$
$\Rightarrow (9^x)((b^2))^x)=16^x$
$\Rightarrow (9b^2)^x=16^x$
$\Rightarrow 9b^2=16$
$\Rightarrow b^2=\frac{16}{9}$
$\Rightarrow b=\pm \frac{4}{3}$
$\Rightarrow (9^x)((9^9)^2)=16^x$
$\Rightarrow (9^x)((b^2))^x)=16^x$
$\Rightarrow (9b^2)^x=16^x$
$\Rightarrow 9b^2=16$
$\Rightarrow b^2=\frac{16}{9}$
$\Rightarrow b=\pm \frac{4}{3}$
- Fri Feb 04, 2011 6:05 pm
- Forum: H. Secondary: Solved
- Topic: Rangpur Higher Secondary 2011/3
- Replies: 6
- Views: 10986
Re: Rangpur Higher Secondary 2011/3
It's two ranges of $f$ and $g$ .It doesn't mean that if $f$ is $-3$ then $g$ is $-2$.leonardo shawon wrote:$-3<f<1$ and $-2<g<4$
if $f=-2.90$ and $g=-1.90$ their multiple $fg$ will give a positive answer!!!
thats why $6<fg<4$