• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Explanation of answer for Cambridge 3U Year 12 book- 3D trig (1 Viewer)

totallybord

Member
Joined
Mar 13, 2006
Messages
209
Gender
Female
HSC
2008
Hi,

Just trying to explain this to cousin but cannot.

The Question is in 2G question 14 part 2 of a. The diagonal PQ of the rectangular prism in the diagram makes angles of alpha, beta and gamma respectively with the edges PA, PB and PC.
(diagram is rectangular prism with base labelled from front left A, P and B and top face Q in the back left corner and C in the front right corner. No other letters are given)
Prove that cos^2alpha+cos^2beta+cos^2gamma=1.
i)This has been proven.
ii) What is the 2 dimensional version of this result? This is the question that I didn't understand. It says in the answers that cos^2alpha+cos^2beta=1 where alpha +beta=90degrees

Please explain.

Thanks!
 

He-Mann

Vexed?
Joined
Sep 18, 2016
Messages
278
Location
Antartica
Gender
Male
HSC
N/A
To go from 3D to 2D, 'squish' the rectangular prism so the result becomes a rectangle. Like, the point C goes to point P.

Refer to the following image: http://pasteboard.co/2cmMy1orE.png

I have added extra labelling. By trigonometry,

AP = rcos(alpha)
AC = rsin(alpha)

BP = rcos(beta)
BC = rsin(beta)

Use Pythag, BP^2 + BC^2 = r^2. However, we know that BC = AP by property of rectangles (opp. sides are same).

This becomes, BP^2 + AP^2 = r^2

(rcos(beta))^2 + (rcos(alpha))^2 = r^2

(cos(beta))^2 + (cos(alpha))^2 = 1

and clearly, alpha + beta = 90 degrees celcius.
 

totallybord

Member
Joined
Mar 13, 2006
Messages
209
Gender
Female
HSC
2008
Hi,
thanks for taking the time but I don't understand your diagram. I'm not sure whether I have been clear with the diagram in the question - I'll post the diagram from the book and if your solution is still the same- let me know. =) Thanks!See the link here labelled trig

trig.png
 

He-Mann

Vexed?
Joined
Sep 18, 2016
Messages
278
Location
Antartica
Gender
Male
HSC
N/A
The solution is the same, my diagram is the base of the rectangular prism.

What don't you understand about my diagram?
 

totallybord

Member
Joined
Mar 13, 2006
Messages
209
Gender
Female
HSC
2008
But in your diagram, APBC is a rectangle whereas in the original diagram, C is directly above P so "squishing" C to P will mean there will not be a rectangle?
 

davidgoes4wce

Well-Known Member
Joined
Jun 29, 2014
Messages
1,877
Location
Sydney, New South Wales
Gender
Male
HSC
N/A
For the first part of this question part (i)

How did you guys go about proving that cos^2 a+cos^2 b +cos^2 y =1


SPent 45 minutes getting close to it but no avail. I tried to break it up as 2 x 2 Dimensional shapes.

Tried something along the lines of:





 

He-Mann

Vexed?
Joined
Sep 18, 2016
Messages
278
Location
Antartica
Gender
Male
HSC
N/A
You labelled your angles wrong. Alpha is the between the diagonal PQ and PA, not the angle between PD and PA as PD is the diagonal of the base of the rectangular prism. Same with beta.

After that, it should be easy.
 
Last edited:

davidgoes4wce

Well-Known Member
Joined
Jun 29, 2014
Messages
1,877
Location
Sydney, New South Wales
Gender
Male
HSC
N/A
You labelled your angles wrong. Alpha is the between the diagonal PQ and PA, not the angle between PD and PA as PD is the diagonal of the base of the rectangular prism. Same with beta.

After that, it should be easy.
That term you labelled in your diagram, has thrown me off in your explanations.

Isn't the angle between PA the same as PQ? Which is the same angle between the planes PA and PD ?



































As you guys can probably see it does have all the cos notations but I can't seem to eliminate PQ from both the LHS and RHS.
 

He-Mann

Vexed?
Joined
Sep 18, 2016
Messages
278
Location
Antartica
Gender
Male
HSC
N/A
Part (ii) seems like the more easier version of this question than part (i) Doing my fcuken head in
Oh, you are right, they are the same. My initial attempt put me in the same predicament you are in.

However, consider the triangles PQA, PQB and PQD.
 

davidgoes4wce

Well-Known Member
Joined
Jun 29, 2014
Messages
1,877
Location
Sydney, New South Wales
Gender
Male
HSC
N/A
Oh, you are right, they are the same. My initial attempt put me in the same predicament you are in.

However, consider the triangles PQA, PQB and PQD.
Thinking it might involve the use of the cosine rule. I spent another 25 minutes this morning on this question and it seems like I am going in circles at the moment.
 

davidgoes4wce

Well-Known Member
Joined
Jun 29, 2014
Messages
1,877
Location
Sydney, New South Wales
Gender
Male
HSC
N/A
Im still along the lines of thinking that

PA^2+PB^2+PC^2=PQ^2 which is what we have to prove.


The problem is if you treat PD as 'r'. I can't seem to find a way in which to get PC in terms of cos^2 y.
 

He-Mann

Vexed?
Joined
Sep 18, 2016
Messages
278
Location
Antartica
Gender
Male
HSC
N/A
Im still along the lines of thinking that

PA^2+PB^2+PC^2=PQ^2 which is what we have to prove.


The problem is if you treat PD as 'r'. I can't seem to find a way in which to get PC in terms of cos^2 y.
If you consider the triangles I mentioned, the answer falls out immediately.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top