Angle Properties of Circles

Given that O is the centre of the circle and ∠OBC = 20⁰, calculate ∠BAC.

Answer explanation

ΔBOC is isosceles triangle.

∠BOC = 180 - 20 - 20

= 140

∠BOC = 2∠BAC

∠ at circumference = 2 × ∠ at centre of the same segment

∴ ∠BAC = 140 ÷ 2

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CCSS.HSG.C.A.2

In the diagram, O is the centre of the circle, and SP is parallel to RQ. Calculate ∠PSR

Answer explanation

Apply co-interior angles

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CCSS.HSG.C.A.2

In the diagram, O is the centre of the circle, and SP is parallel to RQ.

Calculate ∠OPQ

Answer explanation

∠POQ = 40 × 2

= 80⁰

∠OPQ = (180 - 80) ÷ 2

= 50⁰

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CCSS.HSG.C.A.2

O is the centre of the circle and ABE is a triangle. If the circle cuts BE at C and AE at D, calculate ∠x, ∠y and ∠z.

• x = 25⁰, y = 140⁰

  and z = 15⁰

• x = 30⁰, y = 110⁰

  and z = 40⁰

• x = 40⁰, y = 110⁰

  and z = 30⁰

• x = 15⁰, y = 140⁰

  and z = 25⁰

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CCSS.HSG.C.A.2

In the diagram, P, Q, R and S are points on the circle. PS and QR are produced to meet at A. If ∠QSR = 35⁰, ∠QPA = 70⁰ and ∠RSA = 75⁰, calculate ∠PQR.

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CCSS.HSG.C.A.2

In the diagram, P, Q, R and S are points on the circle.

PS and QR are produced to meet at A.

If ∠QSR = 35⁰, ∠QPA = 70⁰ and ∠RSA = 75⁰,

calculate ∠PAQ.

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CCSS.HSG.C.A.2

Given that BD = DC, ∠ABD = 42⁰ and ∠DBC = 65⁰, calculate ∠BDC

Answer explanation

ΔBDC is isosceles triangle.

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CCSS.HSG.C.A.2

Given that BD = DC, ∠ABD = 42⁰ and ∠DBC = 65⁰, calculate ∠BAC

Answer explanation

∠BAC & ∠BDC are angles from the same segment [butterfly/ ribbon]

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CCSS.HSG.C.A.2

Given that BD = DC, ∠ABD = 42⁰ and ∠DBC = 65⁰,

calculate

∠DAB

Answer explanation

ΔBDC is isosceles triangle.

So ∠BCD = 65⁰

Sum of opposite angles in cyclic quadrilateral is 180⁰

∴ ∠DAB = 180 - 65

Tags

CCSS.HSG.C.A.2

Find the values of c and d.

• c = 30⁰, d = 30⁰

• c = 30⁰, d = 65⁰

• c = 65⁰, d = 65⁰

• c = 65⁰, d = 30⁰

Tags

CCSS.HSG.C.A.2