Circle Geometry, SMB-010
Circle Geometry, SMB-008
In the diagram, \(AC\) is a diameter of the circle centred at \(O\), and \(OA = AB\).
Find the value of \(\theta\). (3 marks)
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Find the size of angles \(x^{\circ}\) and \(y^{\circ}\). (3 marks)
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\(x= 38^{\circ}\)
\(y= 104^{\circ}\)
\(x=38^{\circ}\ \ \text{(angles standing on the same arc)} \)
\(\text{Let}\ \ \theta =\ \text{angle at centre on the same arc} \)
\(\theta = 2 \times 38 = 76^{\circ} \)
\(y^{\circ}\) | \(=180-76\ \ \text{(180° in straight line)}\) | |
\(=104^{\circ} \) |
In the circle with centre at \(O\), \(\angle BAC = 36^{\circ}\).
Find \(\theta\). (2 marks)
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\(\theta = 72^{\circ}\)
\(\text{Property: angle at centre is twice angle on circumference, standing on same arc.}\)
\(\theta= 2 \times 36= 72^{\circ}\)
In the diagram, \(AC\) is a diameter of the circle centred at \(O\), and \(OA = AB\).
Find the value of \(\theta\). (3 marks)
\(\theta = 30^{\circ}\)
\(\angle ABC=90^{\circ}\ \ \text{(angle in semi-circle)}\)
\(OA=OB\ \ \text{(radii)} \)
\( \angle OAB=60^{\circ}\ \ ( \Delta OAB\ \text{is equilateral}) \)
\(\theta\) | \(= 180-(90+60)\ \ (180^{\circ}\ \text{in}\ \Delta) \) | |
\(= 30^{\circ} \) |
In the diagram, the vertices of \(\Delta ABC\) lie on the circle with centre \(O\). The point \(D\) lies on \(BC\) such that \(\Delta ABD\) is isosceles and \(\angle ABC = x\).
Explain why \(\angle AOC = 2x\). (2 marks)
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\(\text{See Worked Solutions}\)
The points `A`, `B` and `C` lie on a circle with centre `O`, as shown in the diagram. The size of `angleAOC` is 100°.
Find the size of `angleABC`, giving reasons. (2 marks)
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`130^@`
The points \(A\), \(B\) and \(C\) lie on a circle with centre \(O\), as shown in the diagram.
The size of \(\angle ACB\) is 40°.
What is the size of \(\angle AOB\)?
\(D\)
\(\angle AOB = 2 \times 40 = 80^{\circ}\)
\(\text{(angles at centre and circumference on arc}\ AB\text{)}\)
\(\Rightarrow D\)
The points `A`, `B` and `P` lie on a circle centred at `O`. The tangents to the circle at `A` and `B` meet at the point `T`, and `/_ATB = theta`.
What is `/_APB` in terms of `theta`?
`B`
`/_ BOA= 2 xx /_ APB`
`text{(angles at centre and circumference on arc}\ AB text{)}`
`/_TAO = /_ TBO = 90^@\ text{(angle between radius and tangent)}`
`:.\ theta + /_BOA` | `= 180^@\ text{(angle sum of quadrilateral}\ TAOB text{)}` |
`theta + 2 xx /_APB` | `= 180^@` |
`2 xx /_APB` | `= 180^@-theta` |
`/_APB` | `= 90^@-theta/2` |
`=> B`