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:
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.
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 and
We are told: .
. . Hence: .
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
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.
For angle AOB,
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.
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.
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.