Omnimaga
General Discussion => Other Discussions => Math and Science => Topic started by: Munchor on May 05, 2011, 03:05:53 pm
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I have this image and I need to find out the height of the triangle.
What I know:
- The triangle is rectangular so that angle above is 90º
- The base of the triangle is 50 ( sqroot(40²+30²) ).
I need to find out height (please without really advanced stuff, if I were in USA, I'd be highschool freshman, so something about that level).
(http://img.removedfromgame.com/imgs/mathsquestion.png)
The triangle can also be divided in two rectangular triangles.
EDIT: Also, no trigonometry. I know how to use it, but can't use it this year.
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Seems like you can split that triangle in half and use the Pythagorean Theorem.
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Seems like you can split that triangle in half and use the Pythagorean Theorem.
No, that can't because he doesnt know the hight.
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Seems like you can split that triangle in half and use the Pythagorean Theorem.
Only seems like cos I really can't. I only have 1 side of each triangle (both splitted).
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Ok if you split the triangle the hypotenuse is 30 and the bottom is 25. You can use the Pythagorean Theorem to find missing legs, too.
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Why would the bottom be 25? The triangle's base is not going to be split evenly in half. Before I solve this Scout, you are familiar with what Sin/Cos and the like do?
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30²=x²+y²
40²=(50-x)²+y²
Got it, y is height, thanks much everyone though!
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well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leeds to h = 24, if i'm not mistaken
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well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leeds to h = 24, if i'm not mistaken
I have no idea of what you did Horrowind :S
Solving my system of equations:
30²=x²+y²
40²=(50-x)²+y²
I get x=18 and y=24, which is exactly the answer (24).
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well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leeds to h = 24, if i'm not mistaken
That is an elegant solution! Here I am in my corner stuck on using Sin and Cos :P
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@Builderbot: I tried trigonometry but I couldn't do it because I don't know how to find out the two lower angles. Did you make it? I'd like to know a sin/cos solution too.
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well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leeds to h = 24, if i'm not mistaken
Mind explaining? I'm curious to know why the area is the same as the height :D
Edit: Nevermind, got it!
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well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leeds to h = 24, if i'm not mistaken
Mind explaining? I'm curious to know why the area is the same as the height :D
It isn't, the area can be calculated by:
(b*h)/2 ;b=50 and h=24
(50*24)/2
A = 600
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Yes I just noticed, I read it wrong. Nice solution!
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So i made a new picture to name the angles, there is angle A,B,C and D. We first need to find angle B. Sin = opposite over the hypotenuse. In triangle ABC, the hypotenuse is 50, and the opposite of angle B is 40. So Sin(B)=40/50.
Now we also have triangle ABD, where 30 is the hypotenuse, and side Y is the opposite to angle B. So in this case, Sin(B) = Y/30.
Now we substitute both equations for Sin(B) and get 40/50 = Y/30. Solving yields 24.
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n64_super_mario_64_start.jpg O.O
Very well :)
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Hah whenever I need a fresh image I always find one I'm not using on my desktop and clear it in paint and reuse it :P I don't always remember to rename it too ;D
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Hah whenever I need a fresh image I always find one I'm not using on my desktop and clear it in paint and reuse it :P I don't always remember to rename it too ;D
I imagine your desktop :P
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Its actually almost empty because I just got this computer O.O
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Nice, I like the same-area solution :D
Another way is this: (using the labels in Builder's picture)
ΔABC and ΔDBA are similar triangles, so AC/AD=BC/AB.
So AD=AC*AB/BC=40*30/50=24, as the others got. :)
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well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leeds to h = 24, if i'm not mistaken
That really only works if the triangle is symmetric across both sides of the "height" line. It's not a general formula for the height of a triangle, especially when the triangle is obtuse (not the case here).
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It doesn't have to be symmetric, it just has to be a right triangle, AFAICT, which is the fact that my similar-triangles solution relies upon.
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Er, yeah. I forgot that the right angle is a degenerate case that forces the height into discrete values.
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Oh okay :P We all make mistakes and forget, don't worry. ;D
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well, were is no need for advanced sin/cos-stuff... one can simply use the area of the triangle:
A = (50 * h)/2 = 30*40/2
which leads to h = 24, if i'm not mistaken
And here I was, about jump straight in with law of cosine...
Thumbs up for simplicity!
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What about me? Or are similar triangles too much? :P
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Wait. How are they similar?
(I'm pretty sure you're going to jump in with an obvious answer that I overlooked...)
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By AA (angle-angle). They have a right angle and angle B in common.
Edit: angle B, not A. Fixed.
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Ohhh...
Okay then, a thumbs-up for you too.
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I finally figured it out today. My mind was to busy yesterday I guess.
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you could simply use coordianate(bash) -or to make it sound better analytical geo.
let a=0,0 b=30,0 c=0,40.
the equation of bc is y-40=-4x/3
the distance (using the point to line formula) from 0,0 is just 24!!!!!!(factorial?)
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What the...?
EDIT: I can vaguely understand that post after spending a few minutes trying to decipher it.
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What the...?
EDIT: I can vaguely understand that post after spending a few minutes trying to decipher it.
http://www.intmath.com/plane-analytic-geometry/perpendicular-distance-point-line.php
that problem is actually sort of trivial .........no offense guys.
i don't know but the area olution was a very standard method if you seen it before
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What the...?
EDIT: I can vaguely understand that post after spending a few minutes trying to decipher it.
http://www.intmath.com/plane-analytic-geometry/perpendicular-distance-point-line.php
that problem is actually sort of trivial .........no offense guys.
i don't know but the area olution was a very standard method if you seen it before
Ummm...he's not using coordinates.
When you take the perpendicular from the base to the opposite corner, do you get two similar triangles?
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Hm I am actually not bad with trigonometry. You can solve it with sinus this way:
∠α = tan(40/30) ≈53.13°
sin(α) = sin(53.13°) =0.8
30*0.8 = 24 (since sin gives the ratio of the opposite side to the hypotenuse)