Circle Theorms (2 Viewers)

[davidbarnes]

Can anyone help me with the below circle theorms. I've attempted some, and would appreicate knowing whether they are right or not. As for the rest, I have absolutely no idea how to go about them.

Possible Answers:
1. x = 86 degrees
2. x = 45 degrees
5. x = 27 degrees
6. x = 25 degrees
7. x = 47 degrees

 

[davidbarnes]

Also can anyone please explain the theorm "angles subtened by the same arc" as I just can't seem to get this one no matter what way I look at it.

 

[twilight1412]

refer to the diagram

the angle a & b are equal because they both subtend the same are zw
subtend means to go from a pt to some place basically i could explain it more if you wanted

 

[davidbarnes]

Ok, I clearly get that, although how do I relate it to he #2 problem above?

 

[conman]

I think circle geometry is most easiest part. You only have to learn by heart (I think) 17 theorm and then you be able to solve any questions about circle geometry!!

 

[Riviet]

For 1, 2, 3, and 4, use the "angle at the centre is twice the angle at the circumference standing on the same arc" theorem (angle at centre theorem) to find x.

5: Use the "angles subtended by the same arc" theorem (angles in the same segment theorem) to show that ^PQS is 37 degrees. ^QTR must be 85 degrees (vertically opposite angles are equal), then you can find ^PTQ by subtracting 85+85 from 360 and dividing by two since ^PTQ=^STR (vertically opposite angles). Finally use angle sum of a triangle to find x.

6: The angle in a semi-circle is a a right angle (i.e ^ACB = 90 degrees). Use this and angle sum of a triangle to find x.

7, 8 and 9 all involve applications of the "angle between a tangent and a chord to the point of contact is equal to the angle in the alternate segment standing on the same chord" theorem (angle in the alternate segment theorem). For example, in 7, this theorem would mean that ^ACT = ^ABC. Similarly in 8, ^TRQ=^RPQ.

In 7, note that triangle ABC is isosceles, meaning that ^BCA = ^BAC.

In 9, note that you can't use the angle in the alternate segment straight away because the angle standing on the chord is not at the circumference, it is instead at the centre. So you need to make a construction of an angle that is standing on chord ML but at the circumference. Then use the angle at the centre theorem so that you can now find x using the alternate segment theorem.

I hope that makes sense.

 

[GaDaMIt]

Ok, I clearly get that, although how do I relate it to he #2 problem above?

For #2 the angle in the centre is clearly not on the circumference, so you can't use the theorum that angles on same arc are equal as they're not both on circumference.

Instead, as Riviet said, you refer to the angle at centre = 2x angle at circumference theorum and apply that

 

[Kujah]

In 9, note that you can't use the angle in the alternate segment straight away because the angle standing on the chord is not at the circumference, it is instead at the centre. So you need to make a construction of an angle that is standing on chord ML but at the circumference. Then use the angle at the centre theorem so that you can now find x using the alternate segment theorem.

Could we use properties of an isosceles triangle and that ^NLM + x+15 =90 (tangent is perpendicular to radius if it is drawn at point of contact) to solve x?

 

[Kujah]

1. x=86 (angle at centre of circle is twice the angle at the circumference subtended by the same arc)
2. x = 83 (angles subtended at the circumference by the same arc are equal)
3. x=62.5(angle at centre of circle is twice the angle at the circumference subtended by the same arc)
4. x= 54 (angle at centre of circle is twice the angle at the circumference subtended by the same arc)
5. x=38 (angles subtended at the circumference by the same arc are equal)
6. x=25 (the angle in a semicircle is a right angle)
7. x=53 (angle in the alternate segment theorem)
8. x=14 (angle in the alternate segment theorem)
9. x=33 (angle in the alternate segment theorem; tangent is perpendicular to radius drawn at P.O.C)

I hope i got all of them right Help correct me if im wrong.

 

[davidbarnes]

Thanks Riviet and Kujah particularly.

My answers were:
1. x = 86 - correct
2. x = 109 - wrong
3. x = 42.5 - wrong
4. x = 54 - correct
5. x = 38 - correct
6 x = 25 - correct
7. x = 53 - correct

I had a bit of trouble with 3 and 7, and got both 2 and 3 wrong. I still have absolutely no idea how to do 8. and 9. So could I please have some help with 2, 3 and 8/9.

 

[twilight1412]

2
bca = 2 boa (angle at circumference is half angle at center)

64 = 2(x - 19)
102 = 2x
x = 51

3 prove congurent trianges OAC & OBC (SAS)
then from that

2x - 15 = 140
2x = 155
x = 77.5

8
2x + 17 = x + 31
x = 14

9
as tangents are always perependicular to the center
90 - ((180 - 96)/2) = x + 15

48 = x + 15
x = 33

 

[Riviet]

Could we use properties of an isosceles triangle and that ^NLM + x+15 =90 (tangent is perpendicular to radius if it is drawn at point of contact) to solve x?

Yep, that works and a nice observation. There is usually more than 1 way to do circle geometry questions.

 

[davidbarnes]

2
bca = 2 boa (angle at circumference is half angle at center)

64 = 2(x - 19)
102 = 2x
x = 51

3 prove congurent trianges OAC & OBC (SAS)
then from that

2x - 15 = 140
2x = 155
x = 77.5

8
2x + 17 = x + 31
x = 14

9
as tangents are always perependicular to the center
90 - ((180 - 96)/2) = x + 15

48 = x + 15
x = 33

With 2.
How did you get 2x and 102?

3. I can't do crongruent triangles. I get how you got angle PRQ = angle RQT.

3. Got tha ione thanks.

8. Can see how you got that.

9. I got 90 ((180 - 96), although why did you then divide that by two? Then I got the rest of it.

 

[twilight1412]

2
the angle at the center is double that at teh circumference therefore

double angle at the center = angle at circumference

2(x-19) = 64

2x - 38 = 64

2x = 102

x = 51

3 does not involve prq or rqt
i think you are refering to question 8 where i used angle of alternate segment
9.
because its an iscoceles triangle
so the other 2 angles are equal
let one angle be x

96 + 2x = 180
2x = 84
x = 42

 

[davidbarnes]

Ok, I now get 1, 2, 3,4, 5, and 6, although not 7, 8, and 9.

 

[pLuvia]

7.
Using the angle in alternate segment theorem, ^ABC=64. And since ABC is an isosceles triangle then the base angles are equal, then use sum of triangle to work out the base angles then since the base angles are equal then x+5=one of the base angles. So solve for x

8.
Same here, use the previous theorem then x+31=2x+17, then solve for x

9.
Ok here, MNL is an isosceles triangle. Then the base angles are equal i.e. ^NML and ^NLM. Find one of the base angles first. Now the radius is perpendicular to the point of contact of the tangent with the circle. Hence ^NLM+(x+15)=90

Then solve for x

 

[davidbarnes]

7.
Using the angle in alternate segment theorem, ^ABC=64. And since ABC is an isosceles triangle then the base angles are equal, then use sum of triangle to work out the base angles then since the base angles are equal then x+5=one of the base angles. So solve for x

8.
Same here, use the previous theorem then x+31=2x17, then solve for x

9.
Ok here, MNL is an isosceles triangle. Then the base angles are equal i.e. ^NML and ^NLM. Find one of the base angles first. Now the radius is perpendicular to the point of contact of the tangent with the circle. Hence ^NLM+(x+15)=90

Then solve for x

For 7, is alternate segment theorem the same as the 'angle between a tangent and chord" theorem? Does x = 53 degrees?

8. X = 7 degrees.

9. I do not get this one at all. How do you even know its an isocoles triangle to begin with?

 

[SoulSearcher]

9. NM and NL are radii of the circle, therefore they are equal.