Trigonometry task?
How can I find out the distance AF?
Please look at the task.
The most I can do is figure out the route from A to E.
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Because the triangles ABF and BCF have the same base 10, the heights must be identical. The point F must therefore be as far away from BC as from AB.
The point F=(Fx,Fy) lies on the straight line AE. This is given by
y = x*tan(55°)
This follows:
Fy = Fx*tan(55°)
Because of the same distances:
10-Fx = Fy
This follows:
10-Fx = Fx*tan(55°)
Fx = 10/(tan(55°) + 1) ~ 4.118365096
Fy = 10 – Fx ~ 5.881634904
###
The following applies to AF:
AF2 = Fx2 + Fy2
AF ~ 7.18
If the triangles are to be identical in area and the sides AB=BC=10cm are fixed, then it is sufficient that the angles in the triangles are equal in size, then the triangles are automatically congruent/have the same surface.
When you sign the line BD and mirror the drawing to it, you will automatically get two congruent triangles ABF and BCF. With the aid of the sine set, the length of the distance AF then falls directly out of the triangle ABF.
Try it again, otherwise there is the fixed paint solution here:
What is fix paint? an AI?
Nene, with “fix” I meant “fast” and with Paint the good old Windows drawing program ^^^
I was just saying that I didn’t want to make this big now, but only what I wrote in Paint.
In the sketch btw has become a small number driver, the length for AF should of course be about 7.18 and not 7.81.