3D geometric problem

[wispy hinge]

a rectangle ABPQ and a regular hexagon ABCDEF are non-coplanar, such that the measure of angle PBC = 90, if the measure of the angle between PF and the plane containing the hexagon is 30, then the measure of the angle between PD and the plane containing the triangle PBC is?

[wild gulch]

what part are you struggling at

[wispy hinge]

in the graph, could you help me with that?

[wild gulch]

can you show me your drawing

[wild gulch]

actually you don't need to draw the exact angles, just do a rough sketch of it

[wild gulch]

nvm there's sth you gotta do first

so if we have triangle ABPQ what can we say about its angles?

[wispy hinge]

Triangle? Or rectangle, in that case its 4 angles are 90 degrees

[wild gulch]

yep

you have perpendicular sides

now you have PBC=90, what does it tell you about the two planes

[zealous vigil]

isnt this basically what it's asking for

wait mb

thats a hexagon

oh wait no it is a hexagon

yea

cause if you put the hexagon on the bottom, it looks like part of a hexagonal prism, right?

because PBC is a right angle

[wispy hinge]

So my graph is okay?

[wild gulch]

yep

[wispy hinge]

which are perpendicular?

[wild gulch]

yes

PB is perpendicular to BC and BA so PB is perpendicular to the whole hexagon

what does that tell you about angle PBF?

[wispy hinge]

Its perpendicular

[wild gulch]

alr

so since PB is perpendicular to the hexagon, where is P's projection on the hexagon

[wispy hinge]

at point B?

[zealous vigil]

can’t this be thought of as part of a hexagonal prism, since angle PBC is a right angle, and quad ABPQ is a rectangle?

[wild gulch]

correct

[zealous vigil]

and since the two bases of a prism are parallel, it can make stuff a bit simpler

?

[wild gulch]

it's not any simpler

also there's a chance that logic can be wrong (tho it's right in this case)

[wispy hinge]

Since PFB is a right triangle with angle 30, FB is like sqrt3 times a constant, being a regular hexagon each angle at each vertex measures 120, then triangles FAB and BCD would be congruent, so x gives me arctangent sqrt3/3

[zealous vigil]

myeah true

[wispy hinge]

Where is my mistake? 🥹

[wild gulch]

you're doing good so far

can you calculate the sides of the hexagon then

[wispy hinge]

Well, I could find the sides based on the constant that I put in the notable triangle of 30, as it is a regular hexagon the measurement of the sides is the same, applying the law of cosines I could find them, but how would that help me?

[wild gulch]

just extract as much info as you can

ik because I've already solved it

[wispy hinge]

really?

[zealous vigil]

isn’t the angle we’re solving for the angle between PD and plane PBC? your x is the angle between PD and the hexagon

[wild gulch]

yep

[wispy hinge]

But when joining from point D to the projection of point B on the plane of the hexagon, wouldn't that angle be the intersection of PD and the plane of triangle PBC?

[wild gulch]

that's wrong

remember how the angle is defined

[zealous vigil]

i got the answer but it isn’t one of the answer choices 😭

[wild gulch]

nah I got the right answer

fairly quickly

[wispy hinge]

Its D?

[wild gulch]

Ye

[zealous vigil]

yea i got the wrong answer lol

[wild gulch]

you found the wrong angle

[wispy hinge]

@wild gulch Is the angle position correct?

[zealous vigil]

oh shoot

oh yeah that makes sense

[wild gulch]

the problem says the angle between PD and (PBC)

[zealous vigil]

im stupid lmao

yea i think D is correct

[wispy hinge]

Could you send me the correct graph for the angle I asked for? I think I could solve it with that.

🙏

[wild gulch]

you shouldnt spoil solutions here

[wispy hinge]

???

[wild gulch]

!nosols

[dim zephyr]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[wild gulch]

@wispy hinge can I continue

[wispy hinge]

Where is the angle you are asking for? That's the only thing I need to graph, please.

[wild gulch]

I can help you find it yourself

[wispy hinge]

Pls

[wild gulch]

lemme refine my argument a little

[wispy hinge]

Ok

[wild gulch]

so look at this graph

if I take a point G on line BC so that DG is perpendicular on BC...

why is DG perpendicular to (PBC)

[wispy hinge]

Basically by extending BC I can draw the perpendicular from point D to the plane containing PBC

[wild gulch]

yep

[wispy hinge]

So the angle in question would be PDG?

[wild gulch]

no

the angle you want to find is the angle between PD and its projection on (PBC)

[wispy hinge]

DPG

[wild gulch]

correct

now you pretty much can solve for the rest

I feel like this is the hardest part of the whole problem

[wispy hinge]

thank you very much Masters @wild gulch and @zealous vigil

[wild gulch]

no need to call us masters

even I haven't gotten into uni

[wispy hinge]

I say this out of respect and admiration, but I understand

[zealous vigil]

i barely helped tho i got confused myself haha

but yea this was a difficult question

[wild gulch]

you got a pretty graph tho

[wispy hinge]

Yea

[wild gulch]

this is mine

[wispy hinge]

thanks to both

[wispy hinge]

🧐

[wild gulch]

pathetically ugly ik

[wispy hinge]

question solved, how do I close the thread?

[zealous vigil]

i mean it isn’t art; it’s math, so as long as it’s discernable

[wild gulch]

.solved

I think

[wild gulch]

type this in

[wispy hinge]

Well, as long as you can analyze and solve the problem, the graph in question is fine.

.solved

[dim zephyr]

Solved

Post marked as solved by @wispy hinge.

Use .unsolved if this was a mistake.

[wild gulch]

no need to close it