ANGLE SUBTENDED BY AN ARC AT THE CENTER OF A CIRCLE

Work out the size of each angle marked with a letter. Give reasons for your answers.

Problem 1 :

Solution :

∠AOC + 230˚ = 360˚

∠AOC = 360 - 230

∠AOC = 130˚

a = 130˚

∠ADC = 1/2 ∠AOC

∠ADC = 1/2(130)

∠ADC = 65˚

b = 65˚

Since ABCD is a cyclic quadrilateral, the opposite angles add upto 180.

∠ABC + ∠ADC = 180˚

∠ABC + 65 = 180˚

∠ABC = 180 - 65

∠ABC = 115˚

c = 115˚

Problem 2 :

Solution :

∠AOC = 360 - 106 = 254˚

2∠ABC = reflex of ∠AOC

Reflex of ∠AOC = 2g

g = reflex of ∠AOC /2

= 254/2

= 127˚

So, the value of g is 127˚.

Problem 3 :

P, Q and R are points on the circumference of a circle, center, O. Angle PRQ = 64˚. SP and SQ are tangents to the circle at the points P and Q respectively.

Work out the size of angle

i) PSQ      ii) PQO       iii) POS      iv) QSO

Solution :

∠POQ = 2 ∠PRQ

∠POQ = 2(64˚)

∠POQ = 128˚

i) PSQ :

∠PSQ + ∠POQ = 180˚

∠PSQ + 128 = 180

∠PSQ = 180 - 126

∠PSQ = 52˚

∠OSP = 52˚/2 ==> 26

In triangle PSQ,

∠SPQ + ∠SQP + ∠PSQ = 180

Here ∠SPQ = ∠SQP (SP and SQ are tangents drawn from the external point, they will have same length).

2∠SPQ + 52 = 180

∠SPQ = 64

ii) PQO :

 ∠PQO = ∠SQO - ∠SQP

 ∠PQO = 90 - 64

 ∠PQO = 26

iii) POS :

∠OPS + ∠OSP + ∠POS = 180

90 + 26 + ∠POS = 180

∠POS = 180 - 116

∠POS = 64

iv) QSO :

 ∠QSO = 1/2 ∠PSQ

= 1/2(52)

 ∠QSO = 26˚

Problem 4 :

P, Q and R are points on the circumference of a circle, center, O. Angle PSQ = 60˚. SP and SQ are tangents to the circle at the points P and Q respectively.

a. Work out the size of angle

i) QPS      ii) PQO      iii) PRQ      iv) POQ

b. what type of triangle is PQS?

c. Given that angle OQR = 10, work out the size of angle OPR.

Solution :

a. Given, SP = SQ

i) QPS :

∠PSQ = 60˚, so ∠QPS = ∠PQS

∠QPS = ∠PQS = 1/2(180 - 60)

∠QPS = 60˚

ii) PQO :

∠PQO = ∠SQO - ∠SQP

= 90˚ - 60˚

∠PQO = 30˚

iii) POQ :

∠POQ = 360˚ - (∠PSQ + ∠SPO + ∠SQO)

= 360˚ - (60 + 90 + 90)

= 360 - 240

∠POQ = 120˚

iv) PRQ :

∠PRQ = 1/2 ∠POQ

= 1/2(120)

∠PRQ = 60˚

b. Given, PSQ = 60˚

SP = SQ

So, PSQ is an equilateral triangle.

c. ∠OQR = 10˚

∠RQP = ∠OQR + ∠OQP

= 10˚ + 30˚ = 40˚

∠RPQ = 180˚ - ∠RQP - ∠QRP

= 180 - 40 - 60 = 80˚

∠QPO = ∠SPO - ∠SPQ

= 180 - 60 = 30˚

∠OPR = ∠RPQ - ∠SPO

∠OPR = 80 - 30

∠OPR = 50˚

Problem 5 :

P, Q, R and S are points on the circumference of a circle, center, O. PT and TR are tangents to the circle. OST is a straight line.

Angle OTR = 38˚

Find the size of the angle:

i) ROT  ii) PQR  iii) SRT  iv) PSO  v) PST

Solution :

i) ROT :

∠TPO = ∠TRO = 90˚

90 + 38 + ∠ROT = 180˚

128 + ∠ROT = 180

∠ROT = 180 - 128

∠ROT = 52˚

ii) PQR :

∠POR = 2 ∠ROT

= 2(52˚)

∠POR = 104˚

∠PQR = 1/2 ∠POR

= 1/2 (104˚)

∠PQR = 52˚

iii) SRT :

OR = OS

∠ORS = ∠OSR

∠RSO = 1/2(∠ORS + ∠RSO)

= 1/2(180 - 52)

∠RSO = 64˚

∠SRT = ∠RSO - ∠OTR

∠SRT = 64˚ - 38˚

∠SRT = 26˚

iv) PSO :

∠PSO = ∠RSO = 64˚

v) PST :

∠PST = ∠RST

= 180˚ - ∠RSO

= 180˚ - 64˚

∠PST = 116˚