Finding Right Triangle Lengths

[Iprogram]

I have a problem that i do not have a solution for that i need a little help with, not even sure it can be done? referencing the drawing below i need to find the Height (C) i know the length of the base is 50 so i was trying to play around with tackling it as a right triangle. then i had two known values 25 and 90degrees. ?doesnt seem to work

 

[lev888]

If this is a 'real life' problem chances are there is no unique solution. Can you draw a different shape with the same constraints?

 

[Dr.Peterson]

I have a problem that i do not have a solution for that i need a little help with, not even sure it can be done? referencing the drawing below i need to find the Height (C) i know the length of the base is 50 so i was trying to play around with tackling it as a right triangle. then i had two known values 25 and 90degrees. ?doesnt seem to work

View attachment 14009

If you knew not only that the base is 50 as shown, but also that the angles at the lower left and lower right are both 25 degrees (or whatever), then you could apply basic trigonometry to either of the two congruent right triangles, with side 25 and angle 25 degrees. But if you don't know any angles other than the right angle, then the height is not determined. You need additional information.

Is there anything more that you know?

 

[Iprogram]

Yes this is actually a real world problem that i need to solve. See the below image i need to know the height of the ? i know that the spacing between A and C is 50 but the concave portion of the ovals is what is causing me a problem. Edit....I do know the height of the oval which i am showing at 20.

 

[Dr.Peterson]

I take it you have three tangent ellipses; must they all have the same dimensions?

 

[Iprogram]

I take it you have three tangent ellipses; must they all have the same dimensions?

Yes all Three Ellipses have the same dimension and they will be "nested"

 

[Dr.Peterson]

The answer turns out to be very simple: the height of the triangle is [MATH]20\cdot\frac{\sqrt{3}}{2} = 17.32[/MATH].

I determined this by first drawing the figure with circles,



where the equilateral triangle makes it clear that height CD is [MATH]\sqrt{3}[/MATH] times the radius BD; and then using the fact that if we compress the entire figure vertically, we get ellipses in the same relationship, with the vertical ratios unchanged:



So the height of the triangle is still [MATH]\sqrt{3}[/MATH] times the vertical semiaxis,10. This is independent of the base of the triangle! So the one fact you omitted in the original question is the one fact needed to answer the question.

 

[Iprogram]

The answer turns out to be very simple: the height of the triangle is [MATH]20\cdot\frac{\sqrt{3}}{2} = 17.32[/MATH].

I determined this by first drawing the figure with circles,

View attachment 14019

where the equilateral triangle makes it clear that height CD is [MATH]\sqrt{3}[/MATH] times the radius BD; and then using the fact that if we compress the entire figure vertically, we get ellipses in the same relationship, with the vertical ratios unchanged:

View attachment 14020

So the height of the triangle is still [MATH]\sqrt{3}[/MATH] times the vertical semiaxis,10. This is independent of the base of the triangle! So the one fact you omitted in the original question is the one fact needed to answer the question.

Thank you! I knew i was missing something