k9sar Posted June 8, 2011 Share Posted June 8, 2011 I was reviewing with my son for his geometry final and he had a question that neither of us could solve. I looked online and did not come up with the answer either. I know it must be a 2-step process... have to find one measurement in order to plug into formula for the right answer... but I can't recall my geometry (it's been a few years). Can anyone out there solve this (I actually know the answer from the key given) and tell me how you got there? circle with intersecting lines within it (not intersecting at center of circle). Measurements given for 2 arc lengths and need to solve for the angle 'x'. Problem is that the formula I know is 1/2 the sum of the 2 opposite arc lengths but you don't have both of them (one measurement is the other arc length, not opposite) anyone??? Link to comment Share on other sites More sharing options...
nunya Posted June 8, 2011 Share Posted June 8, 2011 Figure out the answer and walk into a resterant telling them the problem/solution, see if they give you a meal for knowing it In all seriousness geometry was not one of my higher points... I knew what a circle and square were... And what a bribe was Link to comment Share on other sites More sharing options...
Simon Posted June 8, 2011 Share Posted June 8, 2011 What is the answer? I don't want to post my thoughts if I'm way off! Link to comment Share on other sites More sharing options...
jj big shoe Posted June 8, 2011 Share Posted June 8, 2011 When in doubt, the answer is either "C" or "12". That's what got me through high school. 1 Link to comment Share on other sites More sharing options...
KovemaN Posted June 9, 2011 Share Posted June 9, 2011 With the information given, the answer is a function and not a single number. What answer was given? Link to comment Share on other sites More sharing options...
Precise1 Posted June 9, 2011 Share Posted June 9, 2011 I suspect the same thing. I even flipped through the geometry pages in my Machinery's Handbook #25 and while there were several geometric propositions that applied to some extent, there was nothing that would solve for any values without making assumptions. B Link to comment Share on other sites More sharing options...
2milehi Posted June 9, 2011 Share Posted June 9, 2011 Kinda of the brut force way, but here goes. I'll assume the arc measurements are degrees and there are 360 degrees in a circle. x = 1/2 (y + 67) w = 1/2 (z + 147) y + z + 67 + 147 = 360 2w + 2x = 360 Four equations, four unknows. Use substitution to determine variables. Link to comment Share on other sites More sharing options...
k9sar Posted June 10, 2011 Author Share Posted June 10, 2011 the answer key says 50 degrees. Link to comment Share on other sites More sharing options...
2milehi Posted June 10, 2011 Share Posted June 10, 2011 (edited) 1w + 1x + 0y + 0z = 180 0w + 1x - 1/2y + 0z = 67/2 1w + 0x - 0y - 1/2z = 147/2 0w + 0x + 1y + 1z = 146 Bummer I put the equations into matrix form. From there I put the coefficients into a matrix solver but it came back as an indeterminate. Edited June 10, 2011 by 2milehi Link to comment Share on other sites More sharing options...
KovemaN Posted June 10, 2011 Share Posted June 10, 2011 Kinda of the brut force way, but here goes. If the answer is 50*, that would make y 33* and z 113*. I still don't see how you could arrive at that answer with the information given. Link to comment Share on other sites More sharing options...
RJSquirrel Posted June 12, 2011 Share Posted June 12, 2011 I was reviewing with my son for his geometry final and he had a question that neither of us could solve. I looked online and did not come up with the answer either. I know it must be a 2-step process... have to find one measurement in order to plug into formula for the right answer... but I can't recall my geometry (it's been a few years). Can anyone out there solve this (I actually know the answer from the key given) and tell me how you got there? circle with intersecting lines within it (not intersecting at center of circle). Measurements given for 2 arc lengths and need to solve for the angle 'x'. Problem is that the formula I know is 1/2 the sum of the 2 opposite arc lengths but you don't have both of them (one measurement is the other arc length, not opposite) anyone??? Enjoy! Link to comment Share on other sites More sharing options...
2milehi Posted June 17, 2011 Share Posted June 17, 2011 (edited) There is not enough information given to exactly solve the problem, hence why I was getting the indeterminate with matrix math. Say we set y = 1° a really small arc, then x = 34° and z = 145°. w = 1/2 (147° + 145°) w = 146° Now x + w has to equal 180°, which it does. But angle w is dependent of what arc z is and arc z depends on arc y. So there is an infinite number of solutions Edited June 17, 2011 by 2milehi Link to comment Share on other sites More sharing options...
Tungsten Posted June 18, 2011 Share Posted June 18, 2011 I was reviewing with my son for his geometry final and he had a question that neither of us could solve. I looked online and did not come up with the answer either. I know it must be a 2-step process... have to find one measurement in order to plug into formula for the right answer... but I can't recall my geometry (it's been a few years). Can anyone out there solve this (I actually know the answer from the key given) and tell me how you got there? circle with intersecting lines within it (not intersecting at center of circle). Measurements given for 2 arc lengths and need to solve for the angle 'x'. Problem is that the formula I know is 1/2 the sum of the 2 opposite arc lengths but you don't have both of them (one measurement is the other arc length, not opposite) anyone??? It doesn't matter that the lines don't intersect through the center, the arc length inside where the x is will not change as long as the lines intersect within the circle. Knowing that, both sides with angle x have the arc length of 67. Both of those sides combined gives 67+67=134. Angle 2x has the arc length of 134. I don't know the value for z so screw this. Link to comment Share on other sites More sharing options...
2milehi Posted June 18, 2011 Share Posted June 18, 2011 It doesn't matter that the lines don't intersect through the center, the arc length inside where the x is will not change as long as the lines intersect within the circle. Knowing that, both sides with angle x have the arc length of 67. Both of those sides combined gives 67+67=134. Angle 2x has the arc length of 134. I don't know the value for z so screw this. The arc left of x can have an infinite range of solutions. It can be 67°, it could also be 1° and it could be 50°. So take your pick. Link to comment Share on other sites More sharing options...
Tungsten Posted June 18, 2011 Share Posted June 18, 2011 The arc left of x can have an infinite range of solutions. It can be 67°, it could also be 1° and it could be 50°. So take your pick. It does not matter! As long as the lines intersect within the circle, angle 2x will ALWAYS have the arc length of 134. Arc lengths of angle x will be the same on both sides as long as they cross in the same point on the x-axis. Link to comment Share on other sites More sharing options...
2milehi Posted June 19, 2011 Share Posted June 19, 2011 (edited) That is just not true - 2x does have a range of values. Here are some extreme angles that visually prove that. This is a case where 2x will equal 67° plus a very small arc (0.1°). Now 2x = 67.1° This is a case where 2x will equal 67° plus a very large arc (145.9°). Now 2x = 212.9° I have changed the way the intersected circle looks BUT I have not changed any of the values given. ***When it comes down to it, a person came up with this problem. He drew a circle and two intersecting lines within that circle. Then he probably assigned some arc lengths that add up to 360°, threw in a variable "x" and removed some information to make a problem for 9th graders to solve.*** Edited June 20, 2011 by 2milehi Link to comment Share on other sites More sharing options...
RJSquirrel Posted June 19, 2011 Share Posted June 19, 2011 2milehi and Tungsten: I think you both are falling into a couple of traps in your respective arguments here, based on the assumption that the 147 and 67 are angles, or perhaps erroneously assuming the radius of the circle is 1. These cannot be assumed to be so. 67 and 147 are simply arc-lengths, and without knowing the radius of the circle they lie upon, nothing more can be said about them as the intersection of the lines is within the circle, but not at the center of the circle. While manipulating the radius of the circle or the position within it the lines cross at, there is no doubt a sophisticated differential equation describing x as a function of R and the position of the intersection within the circle which I'd rather not expend the energy to solve. If one assumes x to be a constant, which is pretty likely given this is at a high-school geometry level, it greatly simplifies the problem under consideration. Link to comment Share on other sites More sharing options...
2milehi Posted June 19, 2011 Share Posted June 19, 2011 Looking on the 'net, 67 and 147 have to be in degrees. In order to determine the inside angles (in degrees) of the two intersecting lines the outside circle has to be laid out in degrees. So there is no arc length, only arc measure. Also the OP stated "the measure of an angle formed by two lines that intersect inside a circle is half the sum of the measures of the intercepted arcs". Circle Geometry: Anlge Measures Formed By Intersecting Lines As for a differential equation to solve this, I hardly call this "zero-th" order equation a differential. There are no changing lines/circle with respect to time or position - everything is static. Plus this is 9th grade geometry, not 12th grade calculus. I will stand by my answer that there are an infinite number of solutions within bounds. That is 0° < x < 106.5° Link to comment Share on other sites More sharing options...
Tungsten Posted June 29, 2011 Share Posted June 29, 2011 nevermind that, i hate this problem anyway all math people are sadists i don't go beyond solving radius problems Link to comment Share on other sites More sharing options...
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