- In the figure ∟ACB = 42˚. Find ∟x
Solution:
In the figure ∟AOB = 2∟ACB [angle at the centre = twice the angle at the circumference]
∟AOB = 2 * 42˚ = 84˚
∟AOB = x = 84˚
- In the figure ∟ABC is a circle with centre O and reflex ∟AOB = 250 ˚ find ∟x.
Solution:
In the figure ∟ACB angle at the circumference ∟AOB reflex angle at the centre.
∟ACB = 1/2 reflex ∟AOB
∟ACB = 1/2 * 250 ˚ = 125 ˚
∟x = 250 ˚
- In the fig. AOB is diameter, find ∟C
Solution:
In the given figure AOB is diameter of the circle with centre O.
∟AOB = 180 ˚
∟ACB = 1/2 * ∟AOB
∟ACB = 90 ˚
Angle at the circumference reflex angle at the centre.
In the figure ABCD is a circle with centre O. ∟x = 180 ˚ Find
(i) ∟d
(ii) ∟ y
(iii) ∟a
(iv) ∟b
(v) ∟ b + ∟d
Solution:
(i) ∟ADC angle at the circumference ∟AOC angle at the centre.
∟ADC = 1/2 ∟AOC
∟d = ∟ADC = 1/2*108˚
∟d = 54˚
(ii) reflex ∟AOC = 360˚ – ∟AOC = 360˚ – 108˚ = 252˚
∟y = 252˚
(iii) ∟b = ∟ABC = 1/2 reflex ∟AOC
= 1/2 * 252˚ = 126˚
(iv) ∟a + ∟b = 180˚
∟a + 126˚ = 180˚
∟a = 180˚ – 126˚ = 54˚
(v) ∟b + ∟d = 126˚ + 54˚ = 180˚
(vi) ∟e = 1/2 ∟AOC = 1/2*108˚ = 54˚
∟b + ∟e = 126˚ + 54˚ = 180˚
- In the figure diameter AB and chord QR are produced to meet at P. If ∟QPA = 260˚ , ∟QAR = 360˚.Find ∟x and ∟y
Solution:
∟AOB = 180˚
∟AQB =1/2*180˚
∟AQB = 90˚
∟PAR = ∟BQR = x˚
In the ∆ AQP
∟PAQ + ∟AQP + ∟APQ = 180˚
(36˚+ x) + (90˚ + x + 26˚) =180˚
2∟x – 152˚ = 180˚
2∟x = 180˚ – 152˚ = 28˚
∟x = 14˚
In the ∆ARQ,
∟AQR + ∟QAR + ∟ARQ = 180˚
(90˚+ x) + 36˚+ ∟y = 180˚
90˚+14˚+36˚+∟y = 180˚
140˚ + ∟y = 180˚
∟y = 180˚ – 140˚
∟y = 40˚
- In the fig. ∟CBD = 110˚. Find ∟AOC.
Solution:
∟ABC + ∟DBC = 180˚
∟ABC + 110˚ = 180˚
∟ABC = 180˚ – 110˚ = 70˚
∟AOC = 2∟ABC = 2*70˚ = 140˚
- In the fig ∟ADC = 84˚ and AB = BC Find ∟BDC.
Solution:
ABCD is a cyclic quadrilateral.
∟ABC + ∟ADC = 180˚
∟ABC + 84˚ = 180˚
∟ABC = 180˚ – 84˚ = 96˚
In ∆ABC , AB = BC
∟BAC = ∟BDC
∟BDC = 42˚
- In the fig. AB is diameter ∟BAC = 38˚. Find ∟ADC
Solution:
AOB ia diameter of the circle with centre O.
∟ACB = 1/2 ∟AOB = 1/2*180˚ = 90˚
∟ABC + ∟ACB + ∟BAC = 180˚
∟ABC = 180˚ – 90˚ – 38˚ = 52˚
∟ABC + ∟ADC = 180˚
(ABCD is cyclic quadrilateral)
52˚ + ∟ADC = 180˚
∟ADC = 180˚ – 52˚ = 128˚
- In the fig. ∟QXR = 25˚, ∟QRX = 33˚. Find ∟XYZ and ∟PZQ
Solution:
In ∆QXR: RQ is produced to P
Exterior ∟PQX = ∟Q XR + ∟QRX = 25˚ + 33˚ = 58˚
∟PQX = ∟PYX [angles of same segement]
but ∟PQX = 58˚
∟PYX = 58˚
∟XYZ = 58˚
In the ∆XYZ
∟YXZ + ∟XYZ + ∟XZY = 180˚
25˚ + 58˚ + ∟XZY = 180˚
∟XZY = 180˚ – 25˚ – 58˚ = 97˚
Therefore, ∟ABC = 97˚
- In the fig AFD and BFE are straight lines. Find ∟ACF.
Solution:
In the fig ABCD is cyclic quadrilateral.
∟BCF = 180˚ – 110˚ = 70˚
∟BCA = 23˚
∟ACF = ∟BCF – ∟BCA = 70˚ – 23˚ = 47˚ ………(1)
∟FCD = ∟FED – 180˚
∟FCD = 180˚ – 115˚ = 65˚
∟ACB = ∟AFB but ∟ACB = 23˚
∟AFB = 23˚
∟EFD = ∟ECD [angle in the same segment]
We have ∟ECD = 23˚
∟FCE = ∟FCD – ∟ECD = 65˚ – 23˚ = 42˚
∟ACD = ∟ACF + ∟FCE = 47˚ + 42˚ = 89˚
∟ACD = 89˚
1 thought on “Circles – Exercise 4.4.3 – Class IX[state board]”
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