Pythagoras in 3D

Pythagoras in 3D

Postby Guest » Tue Oct 15, 2019 4:17 pm

I can do question a. I got 25m. But I can't do questions b nor c at all.

Please see pictures for the questions and the diagram.
Attachments
IMG_0277.JPG
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IMG_0278.JPG
IMG_0278.JPG (606.91 KiB) Viewed 379 times
IMG_0276.JPG
IMG_0276.JPG (62.59 KiB) Viewed 379 times
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Re: Pythagoras in 3D

Postby Baltuilhe » Tue Oct 22, 2019 9:48 pm

Good night!

a)
Let's fix the origin of the coordinates in the base of SW mast.
So, calling A the SW mast top and B the NE mast top:
A(0,0,23) and
B(16,12,8), right?

The distance between them:
[tex]d^2=16^2+12^2+(23-8)^2[/tex]
[tex]d^2=256+144+225[/tex]
[tex]d=\sqrt{625}=25[/tex]
So, 25 is right :)

b)
From SW to NE, the 'high' of the middle point is:
[tex]\frac{23+8}{2}=\frac{31}{2}=15,5[/tex]

From SE to NW, the 'high'of the middle point is:
[tex]\frac{28+5}{2}=\frac{33}{2}=16,5[/tex]

So, the sparrow is 16,5-15,5=1,0 below the robin.

c)
Try to see that the figure is two trapezes.
They intercept one to each other 6 m distant from the East edge.
And the distance between each cable, is 3.

A figure to try to help
Captura de Tela 2019-10-22 às 21.48.08.png
Captura de Tela 2019-10-22 às 21.48.08.png (256.74 KiB) Viewed 280 times

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