Can anyone tell me how was calculated H' step-by-step?

[Sergiulzr]

Hello,

I want to find the solution for this equation. How did the teacher reach this result for h' step-by-step? Please see the image below:

Thanks!

 

[tkhunny]

Try solving the Pythagorean Identity for [math]\sin\left(\beta\right)[/math], using the positive square root..

 

[Sergiulzr]

I tried a few times but I don't reach this result.

 

[JeffM]

I tried a few times but I don't reach this result.

I am not getting there either. First, I simplified the variables.

[MATH]\text {Set } x = L + L_1 + 0.5D_{2min},\ y = \{L_1 + 0.5(D_{1min} + D_{2min})\},[/MATH]
[MATH]\text {and } z = j_{1min} + j_{2min}.[/MATH]
[MATH]\therefore sin( \beta ) = \dfrac{h'}{x} \text { and } cos ( \beta ) = \dfrac{y - z}{x} \implies[/MATH]
[MATH]1 = \left ( \dfrac{h'}{x} \right )^2 + \left ( \dfrac{y - z}{x} \right )^2 = \dfrac{(h')^2}{x^2} + \dfrac{(y - z)^2}{x^2} \implies [/MATH]
[MATH]x^2 = (h')^2 + (y - z)^2 \implies (h')^2 = x^2 - (y - z)^2.[/MATH]
Now there seems to be a missing step,

namely that [MATH]h' \ge 0.[/MATH] If so,

[MATH]h' = \sqrt{ x^2 - (y - z)^2} = \sqrt{x^2 - y^2 + 2yz - z^2}.[/MATH]
Now, without additional information, I do not see how that reduces to

[MATH]\dfrac{x}{y} * \sqrt{zy}.[/MATH]