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Comment on Circle with Parallel Lines
Awesome! How amazing are your
I solved almost similar with
< AOD = 160 degrees (straight line is 180 degrees)
since AO =OD (equal radii) so < OAD = < ODA = 10 degrees (sum of all sides of triangle is 180 degrees).
Now < OAD = 10 degrees is alternate angle to < ADC which equals to 10 degrees. (alternate angles are equal).
Now we have 2 angles in triangle that has x. So x +90+10 is 180 degrees, sum of all angles of triangle. So x = 80 degrees. Is this correct?
Perfect!
Perfect!
Hi Mr Hanneson,
I used a more simple and very very quick approach for solving this question.
STEP I:-
Since minor arc BD is same for both < BOD and < BAD and <BOD is central angle then < BAD = 10 degrees
STEP II:-
Since we know that line AB and line CD are parallel and line AD is a transverse from those parallel lines then <BAD = < ADC That is 10 degrees.
STEP III:-
Since line ED is a diameter then any angle joining ED is right angle and assigning F to denote at the angle x we get a triangle FCD.
NOW to find <CFD=X
NOW <FCD + <CFD + <FDC =180
Now <ADC=<FDC=10 Degrees
<ECD=<FCD=90 Degrees
Solving we get
<DFC =<X =80 Degrees......VOILA
Cheers,
ABDUL HANNAN
Voila indeed!!
Voila indeed!!
Great solution, Abdul.
great explanation sir
https://gre.myprepclub.com/forum
I already arrived at A before i checked and saw D. This figures not being drawn to scale phenomena is difficult for me to process. Please kindly help with a detailed explanation
Question link: https:/
Here's my full solution: https://gre.myprepclub.com/forum/the-diameter-of-the-circle-is-3073.html...
You're not the only one who is confused. In many cases, the diagrams actually cause more problems (if you believe they're drawn to scale) :-)
Here's a question where the scale is totally off (but it's a GRE-worthy question): https://gre.myprepclub.com/forum/in-the-triangle-abc-what-is-the-value-o...
Cheers,
Brent
Isn't line ED a transversal
All of the steps you took to
All of the steps you took to conclude that ∠CED = 70° are correct.
However, angle x is not an inscribed angle, since the vertex of angle x does not lie ON the circle. Instead, the vertex is INSIDE the circle, which means we can't apply the circle property that says "two INSCRIBED angles containing the same chord must be equal."
Does that help?
Cheers, Brent
Hi Brent,
At the exam, if we would get a problem as this would you recommend us to draw more figures? For example, I found it easier to see the angles formed by the parallel lines when you removed the other lines from the figure. I guess it is a good practice to do this at the exam too, right?
Otherwise, we might not notice them, especially that my figure was drawn really small
Hi Carla,
Hi Carla,
Yes, I definitely recommend redrawing the figure on your noteboard on test day.
Doing so will make it easier to identify and highlight certain aspects of the diagram.
Cheers,
Brent