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fitzpatrick trigonometry problem (1 Viewer)

5647382910

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its difficult for me to post up the question and hard for me to explain the diagram, so if anyone has the fitzpatrick textbook help would be very much appreciated for question 9 on page 24
I have tried everything and still, i cant figure it out
thanks in advance
 

bored of sc

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An open rectangular tank a units deep and b units wide holds water and is titled so that the base BC makes an angle of theta (@) with the horizontal. When BC is returned to the horizontal show that the depth of the water is (a2cot@)/2b units.

AB = DC = a
BC = b

If you look at the diagram dotted lines represent parallel lines

So using alternate angles the angle opposite AB is @ (or theta, same thing).

Using right-angled triangle trigonometry the side BE (if that makes sense) is given by:

tan@ = a/BE

BE = a/tan@
= acot@

The area of the water at the titled state is the area of triangle ABD which is (note I have been talking about D as the point of the triangle opposite side a which is AE)...
using (base x height)/2 = acot@*a/2 = acot@*a/2 = a2cot@/2

The water needs to be the same height when the tank returns to the horizontal position.

So a2cot@/2 = b*depth (area of old triangle formed by water = area of new rectangle formed by water)
thus depth = a2cot@/2b --- divide both sides by b

Yay! I love maths.
 
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youngminii

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MOTHER F*CKING DAMNIT
I got the a^2cot@/2
But I totally didn't see the tilting back depth thingo
I thought we just had to find the area of the water in the triangle
No wonder ==;; Raarrgh
 

5647382910

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bored of sc said:
An open rectangular tank a units deep and b units wide holds water and is titled so that the base BC makes an angle of theta (@) with the horizontal. When BC is returned to the horizontal show that the depth of the water is (a2cot@)/2b units.

AB = DC = a
BC = b

If you look at the diagram dotted lines represent parallel lines

So using alternate angles the angle opposite AB is @ (or theta, same thing).

Using right-angled triangle trigonometry the side BD (if that makes sense) is given by:

tan@ = a/BD

BD = a/tan@
= acot@

The area of the water at the titled state is the area of triangle ABD which is (note I have been talking about D as the point of the triangle opposite side a which is AB)...
using (base x height)/2 = acot@*a/2 = acot@*a/2 = a2cot@/2

The water needs to be the same height when the tank returns to the horizontal position.

So a2cot@/2 = b*depth (area of old triangle formed by water = area of new rectangle formed by water)
thus depth = a2cot@/2b --- divide both sides by b

Yay! I love maths.
how did u get tan @ = a/BD?
and isnt the area of the triangle in the tilted state the are of triangle ABE where E is a point on BC. (this is what i can see from the diagram)
 
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bored of sc

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5647382910 said:
how did u get tan @ = a/BD?
and isnt the area of the triangle in the tilted state the are of triangle ABE where E is a point on BC. (this is what i can see from the diagram)
It's not the area, it's the relationship between the sides of the triangle to get a value for side BD.

I'm using right-angled triangle trigonometry and remember that alternate angles means @ is also equal to the angle opposite side AC (a).

Now we know that tan@ = opp/adj = AC/BE = a/BE

rearranging we get BE = a/tan@ = acot@
 
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5647382910

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bored of sc said:
It's not the area, it's the relationship between the sides of the triangle to get a value for side BD.

I'm using right-angled triangle trigonometry and remember that alternate angles means @ is also equal to the angle opposite side AC (a).

Now we know that tan@ = opp/adj = AC/BD = a/BD

rearranging we get BD = a/tan@ = acot@
Thanks heaps =)
 

bored of sc

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Sorry for the confusion, I meant my is point D was your point E. So I changed D to E above since that is what I meant.

E is the point formed where the water meets the tank at highest point of edge BC. Basically it is located where it says b.
 

g4geexa

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Hey pyro dude, do you have worked solutions to the entire fitzpatrick book?
 
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pyrodude1031

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yes, but i wont upload the entire book. Not good for you.
 

g4geexa

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I'm curious pyro.. did you buy all the chapters individually???
 
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pyrodude1031

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I am not that dumb to buy.
I photocopied 4U Arn from my school
My frend (who didnt do 4u)'s sister went to WGS, who gave me 4U Fitz.
I photocopied 3U Fitz a frend from barker, but the Geometry, & Perm/Comb chapter was missing, so I wrote it up myself during the study break.
 

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