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

totallybord · Nov 1, 2016

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 · Nov 2, 2016

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 · Nov 2, 2016

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

He-Mann · Nov 2, 2016

The solution is the same, my diagram is the base of the rectangular prism.

What don't you understand about my diagram?

totallybord · Nov 2, 2016

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?

He-Mann · Nov 2, 2016

totallybord said:
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?

Change C to D. Made a mistake in labelling vertices.

davidgoes4wce · Nov 3, 2016

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:

$PD^2+DQ^2=PQ^2$

$(r \ cos \alpha )^2 +(r \ cos \beta)^2 + (r \ tan \gamma)^2 =\frac{r^2}{cos^2 \gamma}$

He-Mann · Nov 3, 2016

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.

davidgoes4wce · Nov 3, 2016

He-Mann said:
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 ?

$PA=r \ cos \alpha $ and $ PD= r$

$I then use the Pythagoras theorem of the base to determine the diagonal length$

$PA^2+AD^2=PD^2$

$PA= r \ cos \alpha$

$AD= r \ sin \alpha$

$PB=r \ cos \beta$

$BD=r \ sin \beta$

$AD=BP$

$PA^2+BP^2 = PD^2$

$r^2 \ cos^{2} \alpha + r^2 \ cos^{2} \beta =PD^2$

$I then equated PC^2+CQ^2=PQ^2$

$$ Given that we know $ QC^2=PD^2$

$PC^2+PD^2=PQ^2$

$PC^2+(r^2 \ cos^{2} \alpha + r^2 \ cos^{2} \beta)=PQ^2 \textcircled{1}$

$$ I know if we project PQ towards the ground, $ PQ \ cos \gamma=PC$

$$ Would it be wrong to substitute that into the equation $ \textcircled{1}$

$PQ^2 cos^{2} \gamma+(r^2 \ cos^{2} \alpha + r^2 \ cos^{2} \beta)=PQ^2$

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.

davidgoes4wce · Nov 3, 2016

For those guys that don't have the book, its Q14 from this screenshot:

davidgoes4wce · Nov 3, 2016

Part (ii) seems like the more easier version of this question than part (i) Doing my fcuken head in

He-Mann · Nov 3, 2016

davidgoes4wce said:
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 · Nov 4, 2016

He-Mann said:
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 · Nov 4, 2016

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.

davidgoes4wce · Nov 4, 2016

So I have

$PA=r \ cos \alpha$

$PB=r \ cos \beta$

$PD= CQ$

$\angle PQD=\angle QPC =\gamma$

$tan \ \gamma=\frac{r}{QD}$

Here is my attempt at finding cos y

$tan \gamma=\frac{QC}{PC}$

$Given QC=PD$

$tan \gamma=\frac{PD}{PC}$

$tan \gamma=\frac{\sqrt{r^2 \ cos^{2} \alpha+r^2 \ cos^{2} \beta}}{PC}$

$PC=\frac{\sqrt{r^2 \ cos^{2} \alpha+r^2 \ cos^{2} \beta}}{tan \gamma}$

My answer might be way off.

He-Mann · Nov 4, 2016

davidgoes4wce said:
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.

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

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