In the figure above (not to scale), if AB, BC, CD and DA are equal chords. Find the angle between any two
adjacent chords.
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Step-by-step explanation:
Given−
ABCDEisapolygoninccribedinacirclewithcentreO
andthediameterasAD.
AFhasbeenjoined.
∠FED=110
o
.
AB=BC=CD.
Tofindout−
(i)∠AEF=?(ii)∠FAB=?
Solution−
WejoinOB&OC.
BetweenΔOAB&ΔOBC
OA=OB=OC(radiiofthesamecircleand
AB=BC.
∴BySSStestΔOAB≅ΔOBC⟹OA=BC.
ButBC=AB.
∴OA=AB=BC.
SoΔOABisanequilateralone.
i.e∠OAB=∠ABO=∠AOB=60
o
.
(Eachangleofanequilateraltriangle=60
o
).
AgainAFEDisacyclicquadrilateral.
∴∠FAD+∠FED=180
o
⟹∠FAD=180
o
−∠FED
=180
o
−110
o
.=70
o
.
So∠FAB=∠FAD+∠OAB
=70
o
+60
o
=130
o
.........(ansii).
Now∠AED=90
o
(angleinasemicircle).
∠AEF=∠FED−∠AED=
∴∠AEF=110
o
−90
o
=20
o
........(ansi)
So∠AEF&∠FABarerespectively
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