Solve Hard Trig Question: Cube <ACB Angle

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SUMMARY

The discussion focuses on calculating the angle in a cube formed by vertices A, B, and C. The angle is determined using trigonometric principles, specifically the cosine function. By defining the length of the cube's edge as "s", the lengths of segments AC and BC are expressed as s√3 and s√2, respectively. The final calculation yields an angle of approximately 35.3 degrees, confirming the answer provided in the textbook.

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hard trig question :(

Homework Statement


A,B, and C are three vertices of a cube, as shown in the diagra,. Find the measure of <ACB, to the nearest tenth of a degree.

this is the diagram of the cube
http://img144.imageshack.us/img144/6396/dsdsddry9.jpg

with no given side and angle how do u solve this problem? the answer in the book states 35.3 degrees. but HOW:confused:
 
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HelloMotto said:

Homework Statement


A,B, and C are three vertices of a cube, as shown in the diagra,. Find the measure of <ACB, to the nearest tenth of a degree.

this is the diagram of the cube
http://img144.imageshack.us/img144/6396/dsdsddry9.jpg

with no given side and angle how do u solve this problem? the answer in the book states 35.3 degrees. but HOW:confused:

Find the distance between B and C, say, in terms of the length of one edge (call this "a" or whatever you want). That should be easy. Now, consider the triangle formed by the line going from C to B, then to A and then back to C. Can you visualize that triangle? You know the length of two sides, so you should be able to find all the angles.
 
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yes I can visualize the triangle, but what do you mean i now the length of two sides? there's no given lengths.
 
The angle is independent of the size of the cube- e.g. if you double all lengths then the angle remains the same.

I would solve this problem using the dot product of two vectors.
 
HelloMotto said:
yes I can visualize the triangle, but what do you mean i now the length of two sides? there's no given lengths.

Like I said, simply give a name to the length of one edge. call this length "a", let's say. Then you may find the distance BC in terms of a. Then, when you find the angle, you will see that the calculation will involve taking the ratio of two lengths so the "a" will cancel out and you can get a number (because when you use a trig function, for example tan theta = opposite/adjacent, there is always a ratio of two lenghts involved). It's not surprising that the answer does not depend on the size of the cube for the reason that christianjb explained.
 
im sorry but I am not understanding your method :(. I labled Length BC "a", but I am not sure how you can get an number angle from variables.
 
If it's simpler- just set the length=1.
 
Assume each side of the cube has length "1". That is, side AB has length 1.
Can you find the length of BC?

Once you know those, tan(ACB)= "opposite side divided by near side".

If you used a general "a" you would find that the "a" cancels out when you do that division.
 


set AC as x, set BC as x(sqrt 2) because it is a square.
tan(ACB) = x/x(sqrt2)
= 1/sqrt2
ACB = tan^-1 (sqrt2/2)
 
  • #10


Let s be the length of the segment from A to B.

The length of the diagonal from A to C =s√3
The length of the segment from B to C = s√2

These three segments form a right triangle ABC, with hypotenuse AC.

We can use trigonometry to find angle ACB. Let angle ACB = x

cosx = s√2 / s √3;
cosx = √2/√3 = √6/3;
x = acos(√6/3) = 35.264.
 
  • #11


This thread is more than 4 years old...
 
  • #12


A double necro lol
 
  • #13


mentallic said:
a double necro lol

lol :d
 

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