# Math Help - Tangents and working angles...

1. ## Tangents and working angles...

I'll try to draw a picture in your minds, I seem to not be able to solve it at all and I need it fast :L

There is a circle. It has a polygon inside of it, a bit shaped like a boomerang. This polygon has 4 edges. 3 of the verts are touching the circle. point A is at the bottom and slightly to the right. B is at the top and less to the right than A. C is to the left. O is the centre. the angle of point C of the polygon (inner) is 75 degrees.

There are also two tangents, diagonal lines extending form point P which is to the right. These touch points A and B (bottom and top).

Work out AOB (inner of O) and a reason (please show me a method and Work In Progress)

Then work out the angle of APB, the inner angle of the two tangents meeting at point P.

Sorry for not being able to scan.

2. Even though I think you have described it well, I think I'll struggle to correctly visualise the diagram. Try reading through this page, however, as the theorems are all there:

Circle Theorems

3. Wow it retains exactly what I need *hurries to write more formulas*

I'll have pages by the end, thank you

4. Hello, Mukilab!

I don't know you mean by "the angle of point C (inner)".
But I'll take a guess.

I'll try to draw a picture in your minds.

There is a circle. It has a polygon inside of it, a bit shaped like a boomerang.
This polygon has 4 edges. 3 of the verts are touching the circle.
Point A is at the bottom and slightly to the right.
B is at the top and less to the right than A.
C is to the left. O is the centre.
The angle of point C of the polygon (inner) is 75°.

There are also two tangents, diagonal lines extending from point P which is to the right.
These touch points A and B (bottom and top).

Work out angle AOB.

Then work out angle APB.
Code:
              * * *
*           *  B
*               o
*              *   *
*       *
C o           *       *   *
*       O *         *     o P
*           *       *   *
*       *
*              *  *
*               o
*           *  A
* * *

Draw chords $CA$ and $CB.$

We are told: . $\angle BCA \,=\,75^o$

. . Hence: . $\text{arc}(AB) \:=\:150^o$

Therefore: . $\boxed{\angle AOB \:=\:150^o}$

In quadrilateral $BPAO:\;\;\angle APB + \angle PAO + \angle AOB + \angle OBP \:=\:360^o$

. . $\angle APB + 90^o + 150^o + 90^o \:=\:360^o \quad\Rightarrow\quad \boxed{\angle APB \:=\:30^o}$

5. The boomerang is made of 2 isosceles triangles,
because the centre of the circle is the "inside" vertex of the boomerang.

The triangles are isosceles due to 2 sides being the radius length.
If a triangle is isosceles, then the angles opposite the sides of equal length
are equal (an isosceles triangle is made of 2 identical back-to-back
right-angle triangles).

All 3 angles in a triangle sum to 180 degrees.
The angles in a semicircle (or the 2 angles on one side of a line) sum to 180 degrees.
The angles in a circle sum to 360 degrees.

All of these geometric truths solve these types of question.

$X+Y=75^o$

$W=180^o-2X$

$V=180^o-2Y$

For angle AOB, $180-W+180-V=360-(W+V)=360-(360-[2X+2Y])$

$=2(X+Y)=2(75^o)=150^o$

A well-known result is...
In a circle, the angle at the centre is twice the angle at the circumference
for 2 angles standing on a common arc.
The common arc here is the arc from A to B on the right, just to the left of P.

Triangles oAP and oBP are identical as they have 2 sides the same length
(green dotted hypotenuse and R) and both have a 90 degree angle due to the tangents.
Therefore they are identical right-angled triangles.

The sum of the angles of 2 triangles is 180+180=360 degrees.
Hence, angle APB=360-(90+90+180-W+180-V) or 360-(180+150)=30 degrees.

6. Edit: I do not understand how it is made of two isosceles triangles.

7. Sorry, yesterday I did not have time to view the circle theorem link, only skim and today now I look over it I have solved my first question. Quite easily actually, I apologise but still thank both of you for taking the consideration to explain such a simplistic method to such a simplistic person.

I know need to know the angle of the place where the tangents meet, P (the inner section). This is neither covered int the circle theorem link nor in neither of those posts. Any further help? Sorry for the inconvenience.

8. Hi Mukilab,

If you move A or B to make an non-symmetrical boomerang,
the two boomerang parts (if you break it along the line oC)
will still be isosceles and the maths still follows.

That is if A, B and C are on the circle circumference and if o is the centre.
You can rotate the points A, B and C around the circumference and the 2 triangles of the boomerang will still be isosceles, though different sizes.

In the sketch, the angles X and Y can vary, but both isosceles triangles will still individually have 2 angles the same size.

The only way the triangles would not be isosceles is if any of A, B, C were off the circumference, or if o was not the centre, or if the shape was not a circle.

This problem covers a number of geometric themes,
we can resolve what you need to understand about it yet.

I will send another sketch later.

9. Originally Posted by Archie Meade
Hi Mukilab,

If you move A or B to make an non-symmetrical boomerang,
the two boomerang parts (if you break it along the line oC)
will still be isosceles and the maths still follows.

That is if A, B and C are on the circle circumference and if o is the centre.
You can rotate the points A, B and C around the circumference and the 2 triangles of the boomerang will still be isosceles, though different sizes.

In the sketch, the angles X and Y can vary, but both isosceles triangles will still individually have 2 angles the same size.

The only way the triangles would not be isosceles is if any of A, B, C were off the circumference, or if o was not the centre, or if the shape was not a circle.

This problem covers a number of geometric themes,
we can resolve what you need to understand about it yet.

I will send another sketch later.

Yes, sorry I have just realised this. Please read my update though.

10. Hi Mukilab,

Angle APB is the angle at P (inner angle).
APB means start at A, go to P and end up at B.
Angle BPA is the same angle.

The attachments show a bit on isosceles triangles, tangents and the angle at P.