So this is the question. There is a cuboid $ABCDEFGH$; $ABCD$ is the horizontal rectangular base and $EFGH$ is the horizontal rectangular top. $AB = 4x$, $AE = 3x$, and $BC = y$. If the angle between the skew lines $BH$ and $AD$ is $\theta$ show that $y \sin \theta - 5x \cos \theta = 0$.
I have attempted this question as follows: Since $BD$ is coplanar to $AD$, the angle $\theta$ is also the angle between $AD$ and $BD$. Then, looking at the right-angled triangle formed by $AB$, $AD$ and $BD$, we get $\sin \theta = 4x/BD$, and $\cos \theta = y/BD$. Making both equal to $BD$ and then rearranging algebraically I get $y \sin \theta - 4x \cos \theta = 0$. Not sure why the question says $5x$ and I am getting $4x$ with everything else correct. Any help please?
Since $AD$ and $BC$ are parallel, the angle between $HB$ and $AD$ is the same as the angle between $HB$ and $BC$. Within the cuboid lies the right-angled triangle $HBC$. Using basic trigonometric ratios, we can see that $\sin \theta = HC/BH$, and therefore $\sin \theta = 5x/BH$. We get the $5x$ using Pythagoras to determine $HC$. Rearranging we get $BH = 5x/\sin \theta$. Now we take cosine and find that $\cos \theta = y/BH$. Rearranging we get $BH = y/\cos \theta$. Therefore we can say that $5x/\sin \theta = y/\cos \theta$. Rearranging the algebra gives us our solution.