# Circles – Exercise 4.4.3 – Class IX[state board]

1. 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˚

1. 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 ˚

1. 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.

1. 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.

∟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˚

1. 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˚

1. In the fig. ∟CBD = 110˚. Find ∟AOC.

Solution:

∟ABC + ∟DBC = 180˚

∟ABC + 110˚ = 180˚

∟ABC  = 180˚ – 110˚ = 70˚

∟AOC = 2∟ABC = 2*70˚  = 140˚

1. In the fig ∟ADC = 84˚ and AB = BC Find ∟BDC.

Solution:

∟ABC + 84˚ = 180˚

∟ABC = 180˚ – 84˚ = 96˚

In ∆ABC , AB = BC

∟BAC = ∟BDC

∟BDC = 42˚

1. 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˚

∟ADC = 180˚ – 52˚ = 128˚

1. 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˚

1. 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˚