5. In Fig. 10.39, A, B, C and D are four points on a
circle. AC and BD intersect at a point E such
that 2 BEC = 130º and Z ECD = 20°. Find
Z BAC
B В
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*Question:
⛄⛄⛄⛄⛄⛄
5. In Fig. 10.39, A, B, C and D are four points on a
circle. AC and BD intersect at a point E such
that 2 BEC = 130º and Z ECD = 20°. Find
Z BAC
B В
*Answer:
✍️✍️✍️✍️✍️
Given−
A,B,C&Darepointsonthecircumferenceofacircle.
AC&BDintersectatE.
AB&CDhavebeenjoined.
∠BEC=130
o
&∠ECD=20
o
.
Tofindout−
∠BAC=?
Solution−
WejoinBC.
∠DEC+∠BEC=180
o
.(linearpair)
∴∠DEC=180
o
−∠BEC=180
o
−130
o
=50
o
.
So,inΔDEC,wehave
∠CDE=180
o
−(∠DEC+∠ECD)=180
o
−(50
o
+20
o
)=110
o
.
(anglesumpropertyoftriangles)
NowthechordBCsubtends∠CDE&∠BACtothe
circumferenceofthegivencircleatD&Arespectively.
∴∠CDE=∠BAC=angle
(sincetheangles,subtendedbyachordofacircletodifferent
pointsofthecircumfereceofthesamecircle,areequal).
Ans−OptionB.
solution
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