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Circle Geometry, SMB-015

 

Find the size of angles \(x^{\circ}\) and \(y^{\circ}\).  (3 marks)   

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\(x= 38^{\circ}\)

\(y= 104^{\circ}\)

Show Worked Solution

\(x=38^{\circ}\ \ \text{(angles standing on the same arc)} \)

\(\text{Let}\ \ \theta =\ \text{angle at centre on the same arc} \)

\(\theta = 2 \times 38 = 76^{\circ} \)

\(y^{\circ}\) \(=180-76\ \ \text{(180° in straight line)}\)  
  \(=104^{\circ} \)  

Filed Under: Circle Geometry Tagged With: num-title-ct-path, smc-4240-10-Angles on arcs

Circle Geometry, SMB-010

In the circle with centre at \(O\), \(\angle BAC = 36^{\circ}\).
 

Find \(\theta\).  (2 marks)   

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\(\theta = 72^{\circ}\)

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\(\text{Property: angle at centre is twice angle on circumference, standing on same arc.}\)

\(\theta= 2 \times 36= 72^{\circ}\)

Filed Under: Circle Geometry Tagged With: num-title-ct-path, smc-4240-10-Angles on arcs

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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\(\theta = 30^{\circ}\)

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\(\angle ABC=90^{\circ}\ \ \text{(angle in semi-circle)}\)

\(OA=OB\ \ \text{(radii)} \)

\( \angle OAB=60^{\circ}\ \ ( \Delta OAB\ \text{is equilateral}) \)

\(\theta\) \(= 180-(90+60)\ \ (180^{\circ}\ \text{in}\ \Delta) \)  
  \(= 30^{\circ} \)  

Filed Under: Circle Geometry Tagged With: num-title-ct-patha, smc-4240-10-Angles on arcs

Circle Geometry, SMB-003

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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\(\text{See Worked Solutions}\)

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\(\text{Angles at circumference and centre are both on arc}\ AC\)

\(\therefore \angle AOC = 2x\)

Filed Under: Circle Geometry Tagged With: num-title-ct-patha, smc-4240-10-Angles on arcs

Plane Geometry, EXT1 2017 HSC 12a

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^@`

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`text(Reflex)\ angleAOC` `= 360-100`
  `= 260^@`
`:. angleABC` `= 1/2 xx 260^@` `text{(angles at centre and on}`
 `text{circumference of arc}\ AC)`
  `= 130^@`  

Filed Under: 2. Plane Geometry EXT1, Circle Geometry Tagged With: Band 3, num-title-ct-path, smc-4240-10-Angles on arcs

Plane Geometry, EXT1 2014 HSC 1 MC

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 \(\angle AOB\)?

  1. \(20^{\circ}\)
  2. \(40^{\circ}\)
  3. \(70^{\circ}\)
  4. \(80^{\circ}\)
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\(D\)

Show Worked Solution

\(\angle AOB = 2 \times 40 = 80^{\circ}\)

\(\text{(angles at centre and circumference on arc}\ AB\text{)}\) 

\(\Rightarrow D\)

Filed Under: 2. Plane Geometry EXT1, Circle Geometry Tagged With: Band 1, num-title-ct-patha, num-title-qs-hsc, smc-4240-10-Angles on arcs

Plane Geometry, EXT1 2012 HSC 10 MC

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`? 

  1. `theta/2`  
  2. `90^@-theta/2`
  3. `theta` 
  4. `180^@-theta` 
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`B`

Show Worked Solution

`/_ 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`

Filed Under: 2. Plane Geometry EXT1, Circle Geometry Tagged With: Band 4, num-title-ct-patha, num-title-qs-hsc, smc-4240-10-Angles on arcs, smc-4240-60-Tangents

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