36no one giving it a try.....
IN A RIGHT TRIANGLE ACB, RIGHT ANGLED AT C ANGLE BISECTORS OF A AND B MEET BC AND AC AT P AND Q RESPECTIIVELY. M AND N ARE FEET OF PERPENDICULARS FROM P AND Q ON AB. FIND ANGLE MCN.
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14 Answers
36i mean yup....
i didn't notiice that....
did u do it by plane geometry...?
49My method:
A+B=90°
ΔACP ≡ ΔAPM
Thus, CP=PM
In cyclic quad. ACPM,
<CPM = 180°-A
thus in isoceles ΔBPM,
<PMC=<PCM=A2
O is the incentre..
thus,
OC is the angle bisector too..
thus, <AC0=<PCO=45°
so, <MCO= 90°-A2=B2
Similarly, it can be proved that, <NCO=A2
Thus, <MCN=A+B2=45°
49hmmm.. Why so? I used it in my proof!
Did I go by longer way when there was a possible shorter route anywhere?
If yes please pointout..!
36yup it is....
but trully typical to post it from
the cell phone...
wt let me prepare a rgh sketch
just try to understand it...
coz my drawing is worst.... :o
49I can never access targetiit from cell....:(
It just shows the first 2 posts!
How do u manage to POST from cell?? :O
36i too had the samee problem
but in the setinngs option enable the mobile view.... in opera mini
failed to uupload the img forrmat is png
last option camera wtt...
49a step shortener will be:
<MCB=A/2
similarly, <NCA=B/2
Thus reqd angle= 90°-(A+B)/2 = 45°
36in ur figure
it is easy to observe that BQ inttersects CN at right angles.
which leads to
LPAC + LACN = 45°
OR, LCNO = 45°
OR, LMCN = 45°
now which was better