Correct option is D 28 0square
Given−
O is the centre of a circle.
Its chords AB & CD intersect at Q.
m (arcAD) =
25 o andm(arcBC) = 31 o.
To find out−
∠BQC=?
Solution−
We join AD, BC & BD.
Also we join AO, DO & BO, CO.
25 o i.e ∠AOD = 25 o.
Similarly
m(arcBC)=
31 o i.e∠BOC=31 o.
We know that the angle subtended by a chord of a circle at its centre is twice the angle subtended by the same chord at its curcumference.
∴∠BDC= 2 ×∠BOC=
1
2 ×31 o =15.5 o and
∠ABD= 2 ×∠BDC= 2
1 1
×25 o =12.5 o.
∴In ΔBQD we have
∠BQD=180 o −(∠ABD+
∠BDC)=180 o −(12.5 o +15.5 o ) = 152 o.
∠BQC=180 o −152 o =28 o
(linearpair).
Ans − Option D.
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Answers & Comments
Correct option is D 28 0square
Given−
O is the centre of a circle.
Its chords AB & CD intersect at Q.
m (arcAD) =
25 o andm(arcBC) = 31 o.
To find out−
∠BQC=?
Solution−
We join AD, BC & BD.
Also we join AO, DO & BO, CO.
m (arcAD) =
25 o i.e ∠AOD = 25 o.
Similarly
m(arcBC)=
31 o i.e∠BOC=31 o.
We know that the angle subtended by a chord of a circle at its centre is twice the angle subtended by the same chord at its curcumference.
∴∠BDC= 2 ×∠BOC=
1
2 ×31 o =15.5 o and
1
∠ABD= 2 ×∠BDC= 2
1 1
×25 o =12.5 o.
∴In ΔBQD we have
∠BQD=180 o −(∠ABD+
∠BDC)=180 o −(12.5 o +15.5 o ) = 152 o.
∠BQC=180 o −152 o =28 o
(linearpair).
Ans − Option D.