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Mariana [72]
3 years ago
11

Find the angle between the diagonal of a cube of side length 15 and the diagonal of one of its faces, so that the two diagonals

have a common vertex. The angle should be measured in radians. (Hint: we may assume that the cube is in the first octant, the origin is one of its vertices, and both diagonals start at the origin.)
Mathematics
1 answer:
Daniel [21]3 years ago
7 0

Answer:

the angle between the diagonal is 0.6155 RADIANS

Step-by-step explanation:

Given the data in the question

as illustrated in the image below;

O is [0,0,0]

B is [15,0,15]

E is [15,15,15]

Now to find angle BOE

vector OB = [15,0,15]

vector OE = [15,15,15]

OB.OE = [15,0,15].[15,15,15] = 15 × 15 + 15 × 15 = 450 = |OB||OE|cos[angleBOE]

|OB| = √(15² + 15²) = √(225 + 225) = √450

|OE| = √(15² + 15² + 15²) = √(225 + 225 + 225) = √675

so

Angle BOE = cos⁻¹ (  450 / ( √450 × √675 )

Angle BOE = cos⁻¹ ( 0.81649 )

Angle BOE = 35.265°

Angle BOE = 0.6155 RADIANS

Therefore, the angle between the diagonal is 0.6155 RADIANS  

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this represents the perimeter of the tabletop because we know the width is x. if you drew the rectangle there wouldn't be 1 width. opposite to the width is another line. same with the length. the length is 4x again there are 2 because in a rectangle there are 2 sets of parallel lines. that is why your equation would be 4x+4x+x+x= whatever the perimeter is- you could simplify this equation and write:
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4 0
3 years ago
If 2a+3b=12 and ab=6 find the value of 8a^3+27b^3
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<span>A) 2a + 3b = 12
B) ab = 6 solving for a
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