Please help with the following trig question

[mazda]

I have the following plate.



I need to find the arc length.

Thanks

 

[mazda]

I dont know how to insert the image please show so u can awnser my question

 

[stapel]

I dont know how to insert the image please show so u can awnser my question

To learn how to post images (strictly-speaking, how to post links to images that you have uploaded somewhere web-viewable), please review the relevant FAQ article.

Eliz.

 

[pka]

Go to the very top of this webpage.
Pull down the ‘Forum Help’ tab.
The first item in its list is “Inserting Images”.

 

[mazda]

This is all the information I have (see sketch)

How do you work out the arc length.

P.S it is not a semi circle.

 

[tkhunny]

Subtraction (62-20) gives a right triangle on one of the cone legs.
If A is the outside angle at the bottom of the cone leg, then tan(A) = 55/42
Subtraction gives the inner angle, 180 - 2*A.
One more step.

 

[mazda]

I can work out the triangle, I have all of the triangle lengths and angles. I just can't work out the arc length as it is not a semi circle.

Please help!!

 

[pka]

To find the length of the elliptical arc one must know the ‘height’.
You have not given that measure.
How high above the rectangle is the ‘high-point’ on the arc?

 

[soroban]

Hello, mazda!

                   * *
                *       *
            D *           * B 20
      * - - -*- - - - - - -*- - - *
      |       \           /:      |
      |        \         / :      |
      |         \     r /  :      |
      |          \     /   :55    |55
      |           \   /    :      |
      |            \ /     :      |
      * - - - - - - * - - -+- - - * 
             62     A  42  C  20

I will assume that $\displaystyle AB$ is the radius $\displaystyle r$ of the circle with center $\displaystyle A.$

In right triangle $\displaystyle ABC$, we have: $\displaystyle AC\,=\,42,\;BC\,=\,55$

Hence: $\displaystyle r\:=\:AB\:=\:\sqrt{42^2\,+\,55^2}\:=\:\sqrt{4789}\:\approx\:69.2$


We have: $\displaystyle \,\tan(\angle BAC)\:=\:\frac{55}{42}\;\;\Rightarrow\;\;\angle BAC\:\approx\:52.63^o$

Then: $\displaystyle \angle A\:=\:180^o\,-\,52.63^o\,-\,52.63^o\:=\:74.74^o$


We find the length of arc $\displaystyle BD$ with this proportion: \(\displaystyle \L\;\frac{\widehat{BD}}{2\pi r} \:=\;\frac{\angle A}{360^o}\)

Hence we have: \(\displaystyle \L\,\widehat{BD}\;= \;\frac{2\pi(69.2)(74.74^o)}{360^o} \;\approx \;90.27\)

 

[pka]

Why do you assume that it is a circular arc?
It does not appear to be.
Usually it is so indicated, if it is.

 

[tkhunny]

I can work out the triangle, I have all of the triangle lengths and angles. I just can't work out the arc length as it is not a semi circle.

Please help!!

I was assuming it is a circular arc. If it is, you have enough information. If it is not circular, answer pka.

 

[mazda]

That is the only information I have. There is no height of the arc, I laso don't think it is an arc as tkhunny sugested.

 

[mazda]

THIS IS EXACTLY AS THE QUESTION IS DRAWN AND WRITTEN.



(Lines EF, EG and FG are construction lines only)

The figure above is part of a valve made from SSteel plate which is to be hindged along the side DC.

The dimensions are as follows(all in mm).
AD = BC = 55 DE = EC = 62 AF = GB = 20
FGH is a sector of a circle, note that the area above the horizontal line AB is NOT a semi-circle.

1. Calculate the length EF and the angle FEG in radians.
2. Calculate the arc lentgh FHG and the area FHGE.

I can do question1 but not question 2.

 

[tkhunny]

Great.

If it is circular, you have enough information. If it is not circular, answer pka.

If you have no more information, you're sunk. One can assume all one wants, but that will be exactly correct only by luck.