CAT 2017 Question | Slot - 2 | Quadrilateral in Circle

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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[@ashwin_6810]

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

[@LuikangRobin]

Damn... That's a good question!!

[@souvikdutta5332]

This can be solved using the concept of a cyclic quadrilateral?

[@luckythebestof97]

Otherwise from your diagram CAD=60 hence BAD=90 or BD is diameter and so on

[@KaraYa1997]

Cant we draw a normal quadrilateral diagram in a circle that looks like square and proceed

[@BullseyeBY]

yes

[@kural1000]

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°

[@alokranjan1092]

AOC=30° HOW...?
SHOW IN FIGURE

[@raghulanand1558]

The answer becomes very obvious after seeing the diagram. But how do we arrive at the diagram ?

[@smritisingh3892]

90 degree?

[@mudassiransari4170]

super like

[@jinblaze8442]

"Such is life" I laughed so hard😂😂😂

[@raghavendrapotluri5861]

90 degrees?

[@souvikdutta5332]

This can be solved using the concept of a cyclic quadrilateral as well?

[@surajsingh-hp6yr]

Yes i used cyclic quad.