3D Trig. (1 Viewer)

[NizDiz]

someone help me with the 3d trig question. thanks

 

[NizDiz]

here it is

 

[AnimeX]

i'll give it a go, do you have the answer?

edit: can't seem to get it.. not sure if I can assume <ABP is 90, if not im not sure what to do.

 

[funnytomato]

I guess the 'hard' bit is trying to sketch the situation
A B and C are sequential points with AB=90 and BC=30
From the angles given, P should be 'between' B and C

Then you can express AP, BP and CP in terms of h=PQ
then use the cosine rule in triangles ABP and CBP
you will get a quadratic in h and then should be able to solve it

 

[AnimeX]

I guess the 'hard' bit is trying to sketch the situation
A B and C are sequential points with AB=90 and BC=30
From the angles given, P should be 'between' B and C

Then you can express AP, BP and CP in terms of h=PQ
then use the cosine rule in triangles ABP and CBP
you will get a quadratic in h and then should be able to solve it

how do you know where p is, like why is it between b and c.

do you have a solution, i've attempted it with no luck :s

 

[funnytomato]

how do you know where p is, like why is it between b and c.

do you have a solution, i've attempted it with no luck :s

you can work out where p is by the angle of elevation from A B and C
i.e. inverse tan(larger value) = larger angle = closer to P
so the points , from closest to furthest are B, C , A
which means, while A B and C being sequential , P lies between B and C
with B closer to P

Hopefully that makes sense

 

[funnytomato]

also just follow my hints (in white at #4)
and then use the trig result cos(180-x)=-cos(x)
to find an equation (which simplifies to a quadratic in h), from which the answer quickly follows