Chords AB and CD of a circle intersect in point Q in the interior of a circle as shown in the figure . If m(arc AD) = 20°, and m(arc BC) = 36°, then find Angle BQC
(3 marks)
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Answer:
Given−
Oisthecentreofacircle.
ItschordsAB&CDintersectatQ.
m(arcAD)=25
o
andm(arcBC)=31
o
.
Tofindout−
∠BQC=?
Solution−
WejoinAD,BC&BD.
AlsowejoinAO,DO&BO,CO.
m(arcAD)=25
o
i.e∠AOD=25
o
.
Similarly
m(arcBC)=31
o
i.e∠BOC=31
o
.
Weknowthattheanglesubtendedbyachord
ofacircleatitscentreistwicetheanglesubtended
bythesamechordatitscurcumference.
∴∠BDC=
2
1
×∠BOC=
2
1
×31
o
=15.5
o
and
∠ABD=
2
1
×∠BDC=
2
1
×25
o
=12.5
o
.
∴InΔBQDwehave
∠BQD=180
o
−(∠ABD+∠BDC)=180
o
−(12.5
o
+15.5
o
)=152
o
.
∠BQC=180
o
−152
o
=28
o
(linearpair).
Ans−OptionD.
solution
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