# Application of partial derivatives

#### WY

Hey,

I have no idea where to start, for this question. I know that I will probably have to use vector and scalar product and use the trig identity tan^2(theta)=sec^2(theta)-1 - and of course partial derivatives

Question:
In order to determine the angle theta which a sloping plane ceiling makes with the horizontal floor, an equilateral traingle of side-length l is drawn on the floor and the height of the ceiling above the three vertices is measured to be a,b and c. Show that:
tan^2(theta) = 4(a^2 + b^2 + c^2 - ab -bc - ac)/3t^2

Thanks in advance for anyone's help!

Related Differential Equations News on Phys.org

#### saltydog

Homework Helper
WY said:
Hey,

I have no idea where to start, for this question. I know that I will probably have to use vector and scalar product and use the trig identity tan^2(theta)=sec^2(theta)-1 - and of course partial derivatives

Question:
In order to determine the angle theta which a sloping plane ceiling makes with the horizontal floor, an equilateral traingle of side-length l is drawn on the floor and the height of the ceiling above the three vertices is measured to be a,b and c. Show that:
tan^2(theta) = 4(a^2 + b^2 + c^2 - ab -bc - ac)/3t^2

Thanks in advance for anyone's help!
WY, I've seen your post elsewhere. Others may be like me: I don't know. However, if it were my problem and I HAD to solve it, I tell you what, you can be sure I'd be making a nice size cardboard, wood, plastic whatever sloping ceiling, nothing big, few feet maybe, draw my triangle on the floor, measure the distances and just start working with it. It's an interesting problem and I think, if no one helps you, that's a good place to start.

#### mustafa

Solution

There is no need of partial derivatives. Here is the solution:

Consider the floor to be the x-y plane and the 3 vertices of the equilateral triangle A(0,0,0), B(t,0,0), C(t/2,(sq. root of 3)t/2,0).

The points in the plane of the ceiling will be
P(0,0,a), Q(t,0,b), R(t/2,(sq. root of 3)t/2,c)

The eqn of the plane of ceiling through these points (after simplificaton) is obtained as
(sq root 3)(a-b)x+(a+b-2c)y+(sq root 3)t(z-a) = 0
Eqn of x-y plane is z = 0

The angle theta between these planes is obtained as
cos(theta) = (sq root 3)t/(sq root of (3(a-b)^2+(a+b-2c)^2+3t^2) )

Using tan^2(theta) = sec^2(theta)-1 and symplifying we get
tan^2(theta) = 4(a^2+b^2+c^2-ab-bc-ca)/3t^2

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving