CAT 2017 Question | Slot - 2 | Quadrilateral in Circle
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- Опубликовано: 23 окт 2024
- ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees, then the value of ∠BCD (in degrees) is
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Angle subtended by a chord at centre is double of angle at circumferencd
Hence CAD = 60 ( curcumference )
BAC = 30 given
Hence BAD = 90
Cyclic quad property ~ BCE = 90
Damn... That's a good question!!
This can be solved using the concept of a cyclic quadrilateral?
Otherwise from your diagram CAD=60 hence BAD=90 or BD is diameter and so on
Cant we draw a normal quadrilateral diagram in a circle that looks like square and proceed
yes
Alternative
ODC = OCD= 30 °(isoceles triangle)
Now AOC = 30
Now DAB+DCB= 180 °(cyclic quad)
Then DAC=OCB
DAB+DCB= 2(OCB) +30°+30°=180°
OCB=60°
AOC=30° HOW...?
SHOW IN FIGURE
The answer becomes very obvious after seeing the diagram. But how do we arrive at the diagram ?
90 degree?
super like
"Such is life" I laughed so hard😂😂😂
90 degrees?
This can be solved using the concept of a cyclic quadrilateral as well?
Yes i used cyclic quad.