How to calculate the offset distance of triangle
Solution 1:
Here is a general figure, where the given data are angle $a$ and width $w$.
with $a+2b=90° \ \iff \ a=90°-2b \ \iff \ b=\frac12(90°-a).$
What you need is the value of the vertical displacement $BC$.
As we have
$$\dfrac{BC}{w}=\tan b$$
we deduce that :
$$BC=w \tan b = w \tan(\frac12(90°-a))$$
The numerical application gives:
$$BC=3.5 \tan(33.61 \underbrace{\pi/180}_{*})\approx 2.32715$$
where (*) accounts for the conversion degrees $\to$ radians, which coincide with your calculation.