AB and CD are two equal chords of a circle with centre O. M and N are mid points of AB and CD respectivley and angle AMN = 110 degree. P is any point on minor arc CD. What is the measurement of angle APB? I assure u that the angle will not vary but a constant.
2006-08-18 16:15:25 · 12 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
This is a very nice geometry problem.
I have used Geometer's Sketchpad to to create a graphic of the situation and the answer to the question is almost certainly 70 degrees. However, I am unable to produce a straightforward proof that the answer is 70 degrees.
Note that the result is independent of P because measure of angle APB equals 1/2 of measure of minor arc AB.
Let me know if you want the Geometer's Sketchpad file, or a PDF, or JPEG of the sketch. It is a neat problem.
2006-08-19 04:09:53 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I hope I get the right picture from your description.. If so, my conclusion suggests that the angle is varying depending upon the relative sizes of the circle and the chords..
Let ABCD are 4 distinct consecutive points on the circumference of a circle.
I found AC // BD, i.e. ABCD is a trapezium with vertices at the circumference of a circle (see below).
Let angle ADB = x and angle BDC = y,
Thus, x + y = 70
However, if points B and C are the same point, then y = 0 and x = 70.
What if point C is before point B, i.e. the order of the points on the circumference became ACBD, y < 0 and x >70.
It seems to me that the relative sizes of the chord and circle matter but they were not provided.
If we fixed the size of the circle, we can have short chords AB and CD such that the order of points is ABCD, then angle APB > 70, if points B and C are together, angle APB = 70, and if the order is ACBD, angle APB<70.
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Given AMN = 110
Since AB and CD are equal chord, they are sort of interchangeable. Thus, DNM = AMN = 110.
Consider angles BAD and CDA:
BAD = BAC + CAD and CDA = CDB + BDA
Consider chord BC: BAC = BDC = CDB
Since AB and CD are equal chord, angle BDA = CAD
Thus, angle CDA = BAC + CAD = BAD
Consider the quadrilateral AMND:
Sum of angles = 360
MAD + ADN + DNM + AMN = 360
BAD + ADC + 110 + 110 = 360
2xBAD = 360 - 220 = 140
BAD = CDA = 70
(Thus, AMND is a trapezium because AMN+MAD = 180.)
Consider the quadrilateral ABCD:
Since all 4 vertices are at the circumference of a circle (is it called cyclicquadrilateral),
sum of opposite angles = 180
ABC + CDA = 180
ABC + 70 = 180
ABC = 110
Similarly, BCD = 110
(Thus, ABCD and MBCN are trapezium because
BAD+ABC = 180, and BMN+MBC = 180.)
2006-08-25 07:45:23 · answer #2 · answered by back2nature 4 · 0⤊ 0⤋
70
2006-08-25 04:33:39 · answer #3 · answered by sportsy_chick 2 · 0⤊ 0⤋
70
2006-08-18 17:51:35 · answer #4 · answered by Cor 3 · 0⤊ 0⤋
amn=apb 110
2006-08-24 18:14:16 · answer #5 · answered by stupidgirl 2 · 0⤊ 0⤋
Its not possible. As P is not a constant point. It may be possible if CD the chord is a point actually.
2006-08-25 21:54:29 · answer #6 · answered by sharanan 2 · 0⤊ 0⤋
70 degree
2006-08-22 21:55:25 · answer #7 · answered by lee_axil 1 · 0⤊ 0⤋
If you could have drawn me a picture, I could've figured it out.
2006-08-18 16:20:53 · answer #8 · answered by Carol R 7 · 0⤊ 1⤋
just pay attention in your math classes and read and do well and you will know your angles.
2006-08-18 16:22:05 · answer #9 · answered by mamas_grandmasboy06 6 · 0⤊ 1⤋
wow you people are just too good at math
2006-08-25 16:22:09 · answer #10 · answered by boreddddtodeath 2 · 0⤊ 0⤋