This problem I'm reviewing from a mathematics text for game programming. It involves some trig, and suddenly I realize I'm having difficulty coming to the same conclusion for solving for $k_0$.
From my understanding $k_1$ is intuitive as the relationship of $sin\omega = \frac{o}{h} = \frac{sint\omega}{k_1}$ following some basic algebra to solve for $k_1$
But perhaps this is where I've misunderstood the relationship here, because although the text states a "similar" technique to solve for $k_0$, I'm failing to visualize the triangle in which $k_0$ forms a similar sine relationship to solve for. Personally my mind jumps to cosine usage instead of sine here, but hoping someone can help clear the confusion. Here is the problem from the text:
I have added two purple line segments to your figure, each perpendicular to the line labeled $v_1$.
Since the line labeled $v_1$ is parallel to the line labeled $k_1v_1,$ the two new line segments are congruent. One has length $k_0 \sin \omega$ and the other has length $\sin ((1-t)\omega).$
But another argument is a simple symmetry argument. If we just flip the diagram over so that $v_0$ and $v_1$ exchange places, then $v_t$ is still the sum of two vectors $k_0v_0$ and $k_1v_1$ but now whatever reasoning applied to $k_1v_1$ applies to $k_0v_0$ instead, and the angle between $v_t$ and the horizontal line (now $v_1$) is $(1-t)\omega$ instead of $\omega$.