Help me please.

sin α * R = x (the side shared by both triangles)

if you haven't done trig functions yet, it should be possible to solve with pythagoras though, since it's isoceles. just imagine the hight above the unknown side which splits the upper triangle it into two mirrored right-angled triangles with angles 90°, α/2 and 180-90-α/2 and with hypothenuse R, height h and x/2 as sides.

but something's fishy, i think you can't really solve the problem. you would need two sides to apply pythagoras on the lower triangle. otherwise there are infinite variants of right-angled triangles. every point on a half-circle above a line (center = mid point, radius = half the length) completes into a right-angled triangle. some more angles would also help, but you need *something* more...

>>39683
as >>39694 said you can't solve it with the informations given.

If you draw a circle with diameter = x (the common side) then every point on the circle gives you a right triangle with the end points of the common side x.

There is some missing information.

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I need the height from B to ground level

>>39733
okay, that should help. the ground is a tangent to the circle in point A.

>>39694
>sin α * R = x
I doubt that
it is R * sqrt( 2 - 2cosa )
>law of cosines

>>39683
OP here, i figure it out, i forgot that R is Perpendicular on the ground. the height=2R sin^2(alfa/2)

i have one more problem, my teacher solved it like this: h=R(1-cos(alfa)). so i need a formula from h=2R sin^2(alfa/2) to that.

>>39745
>>39749
ok my h is your g.

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>>39683
faggot op

>>39756
thanks anon.

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>>39760
your welcome

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>>39762
Damn, I got your/you're mixed up. Here's a cutie to fix my mistake

Let M be the center of the circle with radius R.
Draw a perpendicular line from B to AM and you see that the radius R is subdivided into two parts: The height h and a segment which has length R cos alpha .

=> h = R - R cos alpha = R(1 - cos alpha)

>>39774
Thanks a lot! It's easier than this:
>>39756
>>39762