Please help me. I'm curious how to solve it.
- Given the cube $ABCD.EFGH$. If $P$ is the point of extension of $HG$, so that $HP:HG=3:2$, then the angle tangent between the lines $AP$ and $HB$ is $\ldots$
(a) $3 \sqrt{2}\quad$ (b) $\dfrac{7\sqrt{2}}{2}\quad$ (c) $4\sqrt{2}\quad$ (d) $\dfrac{9\sqrt{2}}{2}\quad$ (e)$5\sqrt{2}$
- Given the cuboid $ABCD.EFGH$ has $AB=3,$ $BC =2$, dan $AE = \sqrt{12}$. cosine of the angle between the diagonals $AG$ and $HB$ is $\ldots$
(a) $\dfrac{9}{25}\quad$ (b) $\dfrac{8}{25}\quad$ (c) $\dfrac{7}{25}\quad$ (d) $\dfrac{6}{25}\quad$ (e) $\dfrac{4}{25}$
To number 1:
Look at triangle HPI where I is the intersection. We have that $tan(HPI)=tan(HPA)=2\sqrt(2)/3$ Last equality holds because if $HG=1$ then $HA=\sqrt(2), HP=3/2$.
We also know $tan(IHP)=tan(BHG)=\sqrt(2)$
Now try to use this identity https://math.stackexchange.com/a/2642511/692896
To number two:
Let $I$ be the intersection of $AG$ and $HB$. You can calculate the length $HB=AG$. If you have it you get $HB/2=HI=GI$. Now you can you the law of cosine in the triangle $HGI$
Questions are welcome. At number 1 you may get some more calculation to do. I am pretty sure there might be some nicer way to do it at number 1 but number 2 is pretty simple this way. It's my first time seeing the formula in Nmbr1 so don't think that it's a standard trick