In figure, O is centre of the circle and ZAOB = 60° then find LACB
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Answer:
Given O is the centre of circle.
∠AOB=60
To find : ∠ACB, ∠ADB
∠ACB= 1/2 ∠AOB
(Angle subtended by the arcA at centre is twice the angle subtended by the arc on the circle)
∴ ∠ACB=30°
ACBD is a cyclic quadrilateral
We know that,
Opposite angles in a cyclic quadrilateral sum= 180 °
∠ACB+∠ADB=180°
30° +∠ADB=180°
∠ADB=150°
Hence
∠ACB=30°
∠ADB=150°
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Answer:
3+2
2
1
=
3+2
2
1
×
3−2
2
3−2
2
=
(3)
2
−(2
2
)
2
3−2
2
=
9−8
3−2
2
=3−2
2
=3−2(1.4)
=3−2.8
=0.2
3+2
2
1
=
3+2
2
1
×
3−2
2
3−2
2
=
(3)
2
−(2
2
)
2
3−2
2
=
9−8
3−2
2
=3−2
2
=3−2(1.4)
=3−2.8
=0.2
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