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Sia ? 4 years, 9 months ago
A 91,
DEEPMALA SOCIETY,
JAIPUR, RAHASTHAN.
X-MONTH-20XX
TO
THE EDITOR
JAIPUR.
SUBJECT = APPRECIATING INDIAN GOVERNMENT FOR BANNING OF FIRE CRACKERS.
SIR, MAM,
THROUGH THE ESTEEM COLUMN OF YOUR NEWS PAPER I HAVE READ AN ARTICLE ABOUT BANNING OF FIRE CRACKERS. I FELT HAPPY AT THAT TIME AS I WAS THINKING THAT OUS SOCIETY IS GOING TO EXEMPTED FROM THIS POPULATED ENVIRONMENT AND HAZARDUOUS NOICE OF THE CHRACKERS.
I ALSO AGREE WITH THAT BUT THE POLLUTION IS NOT CONTROLLED YET.
JUST LIKE DELHI AND RAJASTHAN WE HAVE TO CHECK ALL PART OF THE COUNTRY AS WELL.NOT ONLY THE FIRE CRACKERS POLLUTE THE ENVIRONMENT BUT THE CARS AND AUTOMOBILE VEHICLES, FACTORIES, ETC ALSO. SO, WE SHOLD HAVE TO CONTROLL THIS BY USING PUBLIC TRANSPORT, CONSERVATION OF NATURAL RECOURCES ETC.
" LIFE IS ONE AND WE SHOUDNT LET THE POLLUTION (WHICH IS OBVIOUSLY SPREADED BY US) RUIN THIS "
I WILL WAIT FOR YOUR PROMPT REPLY
THANKYOU
YOUR SINCERELY,
ROHIT
Posted by Sakshi Kumar 5 years, 2 months ago
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Sia ? 4 years, 9 months ago
Let ABCD be a quadrilateral circumscribing a circle with centre O.
Now join AO, BO, CO, DO.
From the figure, ∠DAO=∠BAO [Since, AB and AD are tangents]
Let ∠DAO=∠BAO=1
Also ∠ABO=∠CBO [Since, BA and BC are tangents]
Let ∠ABO=∠CBO=2
Similarly we take the same way for vertices C and D
Sum of the angles at the centre is 360o
Recall that sum of the angles in quadrilateral, ABCD = 360o
= 2(1+2+3+4)=360o
=1+2+3+4=180o
In ΔAOB, ∠BOA=180−(1+2)
In ΔCOD, ∠COD=180−(3+4)
∠BOA+∠COD=360−(1+2+3+4)
=360o–180o
=180o
Since AB and CD subtend supplementary angles at O.
Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
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Kiran Kumari 5 years, 2 months ago
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