Rate of change of arc wrt fixed external point

[RayM]

Dear all,

Please can I have some help on a problem I have come across. The image describes my issue.
I would like an equation that gives X as a function of theta, x=f(theta), as my point z traverses the arc.
Empirically measuring x versus theta does not give me anything I recognize as a function, so it is more complicated than I can figure out.
I wasn't sure if this question should appear under trig/geometry or calculus, so apologies if I have the wrong place.
Any ideas would be gratefully received.
Thanks

 

[Dr.Peterson]

It appears that you are not asking for a rate of change of anything, but only for distance x as a function of angle theta. So this is not a calculus problem, but trigonometry.

First, let's use exact values. If the leg BC is 75, then hypotenuse AB will be [MATH]\sqrt{75^2 + 75^2} = 75\sqrt{2}[/MATH].

Next,you might consider working with triangle AZB, in which you know two sides and an angle (SSA). How are you at solving oblique triangles?

 

[RayM]

It appears that you are not asking for a rate of change of anything, but only for distance x as a function of angle theta. So this is not a calculus problem, but trigonometry.

First, let's use exact values. If the leg BC is 75, then hypotenuse AB will be [MATH]\sqrt{75^2 + 75^2} = 75\sqrt{2}[/MATH].

Next,you might consider working with triangle AZB, in which you know two sides and an angle (SSA). How are you at solving oblique triangles?

Thank you. Just the right way to look at it and solve the equations - used cosine rule on triangle AZB as suggested. I then find a quadratic in x and solve with the correct solution.

Many thanks again.

 

[Dr.Peterson]

Another approach would be to use the law of sines to find the angles, then use the law of cosines directly to get x. Either way should give the same result; your way makes naturally gives two solutions, while mine will only find the "other" one if you pay attention.