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Please help me. I'm curious how to solve it.

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  1. 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)$5\sqrt{2}$

  1. 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) $\dfrac{4}{25}$

Blue
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Dom Kang
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    Welcome to MSE. Please write your question here instead of linking to an external site. Please write an informative title. – Randy Marsh Jun 11 '23 at 07:44
  • Take coordinates : for example for the first question : $A(0,0,0),B(1,0,0),D(0,1,0),E(0,0,1),P(3/2,1,1),H(0,1,1)$. Then compute dot product in two ways $\vec{AP}.\vec{HB}=\cos(\alpha)|\vec{AP}| |\vec{HB}|=xx'+yy'+zz'$. Once you have the cosine, it is easy to deduce its tangent. – Jean Marie Jun 11 '23 at 08:55
  • wah thank you, but i still confusing :( – Dom Kang Jun 11 '23 at 09:26
  • i don't know how to use that, i never get that way in my school – Dom Kang Jun 11 '23 at 09:29
  • and how we solve it Professor if coordinate and vector not allowed just use non-coordinate geometry? – Dom Kang Jun 11 '23 at 10:20
  • Welcome to Math.SE! ... Please only ask one question per post. Also, the community prefers/expects a question to include something of what the asker knows. (What have you tried? Where did you get stuck? etc) This helps answerers tailor their responses to best serve you, without wasting time (theirs or yours) explaining things you already understand or using techniques (such as vectors) beyond the intended level of the question. (Generally, isolated problem statements tend to be interpreted as do-my-homework-for-me questions, attracting down- and close-votes.) Good luck! – Blue Jun 11 '23 at 10:24
  • Yeah thank you sir for you answer and suggestion, i'm so sorry this is my first time on Math.SE, and this problem from past question university exam entrance in my country – Dom Kang Jun 11 '23 at 10:57
  • There's no official solution and nothing people talk about the problem. Sorry for my English, i'm not native speaker – Dom Kang Jun 11 '23 at 11:06
  • As Blue has said : say us what you know. Which kind of chapter do you study in 3D geometry at present ? Do you know for example cosine's law : $a^2=b^2+c^2-2bc \cos \hat{A}$ ? – Jean Marie Jun 11 '23 at 15:48
  • HINT. $AH=\sqrt2\ l$, $PH={3\over2}l$. – Intelligenti pauca Jun 11 '23 at 17:56

1 Answers1

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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

Csaba Daniel Farkaš
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