h t t p s://ibb.co/Hz898r4
enter link for picture without spaces.
please help. I got the secants, chords and tangents. I think I found two angles out of the 19. I would appreciate if someone helped me get the rest of the angles.
tangent: <cde
secants:
<cbg
<agf
<abd
chord
<bd
<df
<bg
<bf
<gd
<fg
I think the first angle is 45 degrees and the sixth angle is 55 degrees.
I can't figure out the location of the 50°, 100° and 140° angles
I am guessing that they represent the corresponding arclengths?
e.g. arc BG would be 50/360 of the circumference?
Anyway, it is past midnight, if nobody works at it I will look at it again in the morning.
I assumed that 50, 100, and 140 were supposed to add to 360 so for the bottom (between g and f) I added 70 on the outside of the circle. Also, have a good night and thank you.
Ok, based on the fact that the angle subtended by a chord at the centre is twice the angle
subtended by that same chord at the circle, or the angle subtended by an arc is half the angle subtended at the centre.
We have only one circle so the same centre and radius.
So, if we assume for instance, that the central angle subtended by the chord DF = 140°
then angle#4 = 70°
work your way around, since you are given the central angles all around the circle
that is,
#3 = 35°
#4 = 70°
#14 = 50°
#11 = 50° , notice angles subtended by the same chord on a circle are equal
You can thus fill in all the angles inside the quadrilateral BGFD
Now look at #'s 2, 3, and 4. They form a straight line (180°), you have 2 of them
so #2 = 75°
Same for #'s 13, 14, and 15
You can also find the 4 angles #'s 16,17,18, and 19
e.g. #17 = 180-35-50 = 95° , etc
#5 is now easy
For #7, the angle between a tangent and a chord is equal to the angle subtended by that chord at the circle, so
#7 = 50°
which then gives you #6 = 55
Last, but not least #10 = .....
I think we got all the angles
Thankfully we are simply to LIST all the tangent lines, chords and secants
(to calculate their lengths would be a major major problem.
There is an actual relationship between arclength and chord length as seen in this
webpage:
https://www.1728.org/cntlangl.htm
That makes so much more sense. Thank you. My problem was figuring out which had the same measurements, as well as remembering the formulas in order to calculate the lengths. I understand a lot better now.
To help you find the remaining angles, let's analyze the information you have provided:
- Tangent: ∠CDE
- Secants: ∠CBG, ∠AGF, ∠ABD
- Chords: ∠BD, ∠DF, ∠BG, ∠BF, ∠GD, ∠FG
To solve for the angles, we'll need to use the properties of intersecting lines and circles. Here's how to approach each case:
1. Tangent (∠CDE):
Since ∠CDE is a tangent, it forms a right angle with the radius drawn from the point of tangency (E in this case). Therefore, ∠CDE = 90 degrees.
2. Secants (∠CBG, ∠AGF, ∠ABD):
To solve for these angles, we'll use the Intersecting Secants Theorem, which states that when two secant lines intersect outside a circle, the product of their external segments is equal. We can set up equations using this theorem:
a) ∠CBG
The external segment of ∠CBG can be found by multiplying the external part of ∠CBG (i.e., <CBG) with its external part: (∠CBG + ∠BD) * (∠CBG + ∠BF) = (∠GBD + ∠DBG) * (∠GBF + ∠BFG)
Simplifying, we have: (∠CBG + ∠BD) * (∠CBG + ∠BF) = (∠CBG + ∠BD) * (∠ABD + ∠GBD)
Since both external segments are the same, we can set (∠CBG + ∠BD) equal to (∠ABD + ∠GBD) and solve for ∠CBG.
b) ∠AGF
You can follow a similar process to find ∠AGF by setting (∠AGF + ∠BF) * (∠AGF + ∠FG) = (∠FGA + ∠AGF) * (∠GBA + ∠FGB).
c) ∠ABD
To find ∠ABD, you can use the tangent-secant theorem, which states that the measure of an angle formed by a tangent and a secant drawn from the same point outside a circle is equal to half the difference of the intercepted arcs. In this case, ∠ABD will be equal to (½ * (∠FG + ∠GB + ∠DB)) - ∠GBD.
3. Chords (∠BD, ∠DF, ∠BG, ∠BF, ∠GD, ∠FG):
To solve for these angles, we'll use the Intersecting Chords Theorem, which states that when two chords intersect in a circle, the product of the segments of one chord is equal to the product of the segments of the other chord. We can set up equations using this theorem:
a) ∠BD
The segment multiplication equation will be: (∠BG + ∠GD) * (∠BG + ∠FG) = (∠GB + ∠BF) * (∠GB + ∠DF)
b) ∠DF
Similarly, you can find ∠DF by setting (∠BG + ∠GD) * (∠FG + ∠GD) = (∠GB + ∠BF) * (∠BA + ∠GF)
c) ∠BG, ∠BF, ∠GD, ∠FG
To solve for these angles, you'll need additional information about the lengths or angles in the diagram, as the Intersecting Chords Theorem alone cannot determine them.
By using the provided theorems and properties, you can set up and solve the equations to find the missing angles. Good luck!