finding length of line segment of circle given its arc length and height of ends and mid-point of arc

There isn't a solution to your equations.

Are you sure the diagram is correct (for example did you perhaps swap the positions of "E" and "F") or maybe you misread one of the numbers in the original question?
 
Do you mean that had the values 50, 10 and 80 been different then a solution could be obtained mathematically ?
Though my question and points are same as give.
But , if you take some other numbers instead of 10,50 and 80 and provide solution then also it is good as the main point is solving an equation containing angle and its trigonometric ratio.
 
inanisumeet said:
Please see image at https://ibb.co/MRf0pCC

As you can see I am unable to solve equations
(1) 2 r θ = 80
(2) r cos θ + 40 = r

to get value of 2 r sin θ .

Thanks for reading .
Click to expand...
If your equations are correct, then you can "approximate" r and θ by using Newton's method or Taylor series expansion of cosθ .
 
Cubist said:
There isn't a solution to your equations.
Click to expand...

...I think there's no solution because I sub'd (1) into (2) to eliminate θ

r*cos(40/r) + 40 = r

Then I plotted the following to see if it crosses (or touches) the x-axis which would give a solution:-

r*cos(40/r) + 40 - r

...and it stays away from the x-axis.
 
Clearly the arc AG has to be longer than the sagitta EG, so your numbers are wrong. Replace 80 with something large enough (say, 120) and it could be done.

But you'd need either a graphing program or a numerical method, as in #5 and #4 respectively. Equations with x both inside and outside a sine can only be solved exactly in very special cases, if at all.
 
Cubist said:
...I think there's no solution because I sub'd (1) into (2) to eliminate θ

r*cos(40/r) + 40 = r

Then I plotted the following to see if it crosses (or touches) the x-axis which would give a solution:-

r*cos(40/r) + 40 - r

...and it stays away from the x-axis.
Click to expand...
What tool did you use to plot ?
 
You must log in or register to reply here.
Top