CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
362
You visited us 362 times! Enjoying our articles? Unlock Full Access!
Question

In Figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that BEC=130oandECD=20o. Then BAC is
244145_c65202a7308845cd9b38547ee10c2139.png

A
140o.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
110o.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
70o.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
120o.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 110o.
GivenA,B,C&Darepointsonthecircumferenceofacircle.AC&BDintersectatE.AB&CDhavebeenjoined.BEC=130o&ECD=20o.TofindoutBAC=?SolutionWejoinBC.DEC+BEC=180o.(linearpair)DEC=180oBEC=180o130o=50o.So,inΔDEC,wehaveCDE=180o(DEC+ECD)=180o(50o+20o)=110o.(anglesumpropertyoftriangles)NowthechordBCsubtendsCDE&BACtothecircumferenceofthegivencircleatD&Arespectively.CDE=BAC=angle(sincetheangles,subtendedbyachordofacircletodifferentpointsofthecircumfereceofthesamecircle,areequal).AnsOptionB.
302987_244145_ans.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Tango With Straight Lines !!
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon