An acute angle is an angle that is less than 90°. Such angle can be of different shapes but the fact that their angles are far less than 900, it makes them an acute angle

The acute angle between two planes e.g. the two adjacent faces of a polyhedron also known as a dihedral angle, may be used in describing an acute angle that is between two lines which are normal to the planes

Figure A

α2 = α1 + ϕ. Therefore, ϕ = α2 - α1. We will find the value of + from the slopes of the drawn lines L, and L2 in an acute angle, as follows:

- tan ϕ = tan (α2 - α1) = tan α2-tan α11+tan α1 tan α2
- Recalling that the tangent of the acute angle of inclination equals the slope of the line, then we have:
- tan α1 = m1 which stands for the slope of L1 of the acute angle
- tan α2 = m2 which stands for slope of L2 of the acute angle
- such that m2 > m1
- Therefore, substituting the above expression in tangent formula we have: tan ϕ = m2 - m11+ m1 m2

Example: Referring to figure A, find the acute angle that is between the two lines with m, = 12 and m2 = 2 for their slopes

*Solution*:- tan ϕ =2- 12 ÷ 1 + (12)(2) = 34 =.75
- Such that ϕ = arctan (.75) = 360 52’
- If one of the lines in this acute angle was parallel to the Y-axis, then its slope would be infinite and it will render the slope formula for tan + valueless as a result of an infinite value in both the numerator and denominator of the fraction m2 - m11+ m1 m2 produces an intermediate form of an acute angle 1 + m,m2
- However, if only one of the lines of this acute angle is known to be parallel to the Y-axis, the tangent of + can be expressed in another way. Assume that L2 of fig A was parallel to Y-axis. Then we would have something like this; α2 = 900
- But if L1 has a
*positive*slope, then the acute angle that lies between L1 and L2 would be found by; ϕ = 900 – α1 then tan ϕ = cot α1 = 1m1. If L1 of the acute angle has a*negative*slope, then ϕ = 900 – α1 = -(900 - α1) while tan ϕ = - cot α1 = - 1m1

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An acute angle is an angle that is less than 90°. Such angle can be of different shapes but the fact that their angles are far less than 900, it makes them an acute angle

The acute angle between two planes e.g. the two adjacent faces of a polyhedron also known as a dihedral angle, may be used in describing an acute angle that is between two lines which are normal to the planes

Figure A

α2 = α1 + ϕ. Therefore, ϕ = α2 - α1. We will find the value of + from the slopes of the drawn lines L, and L2 in an acute angle, as follows:

- tan ϕ = tan (α2 - α1) = tan α2-tan α11+tan α1 tan α2
- Recalling that the tangent of the acute angle of inclination equals the slope of the line, then we have:
- tan α1 = m1 which stands for the slope of L1 of the acute angle
- tan α2 = m2 which stands for slope of L2 of the acute angle
- such that m2 > m1
- Therefore, substituting the above expression in tangent formula we have: tan ϕ = m2 - m11+ m1 m2

Example: Referring to figure A, find the acute angle that is between the two lines with m, = 12 and m2 = 2 for their slopes

*Solution*:- tan ϕ =2- 12 ÷ 1 + (12)(2) = 34 =.75
- Such that ϕ = arctan (.75) = 360 52’
- If one of the lines in this acute angle was parallel to the Y-axis, then its slope would be infinite and it will render the slope formula for tan + valueless as a result of an infinite value in both the numerator and denominator of the fraction m2 - m11+ m1 m2 produces an intermediate form of an acute angle 1 + m,m2
- However, if only one of the lines of this acute angle is known to be parallel to the Y-axis, the tangent of + can be expressed in another way. Assume that L2 of fig A was parallel to Y-axis. Then we would have something like this; α2 = 900
- But if L1 has a
*positive*slope, then the acute angle that lies between L1 and L2 would be found by; ϕ = 900 – α1 then tan ϕ = cot α1 = 1m1. If L1 of the acute angle has a*negative*slope, then ϕ = 900 – α1 = -(900 - α1) while tan ϕ = - cot α1 = - 1m1

An acute angle is an angle that is less than 90°. Such angle can be of different shapes but the fact that their angles are far less than 900, it makes them an acute angle

The acute angle between two planes e.g. the two adjacent faces of a polyhedron also known as a dihedral angle, may be used in describing an acute angle that is between two lines which are normal to the planes

Figure A

α2 = α1 + ϕ. Therefore, ϕ = α2 - α1. We will find the value of + from the slopes of the drawn lines L, and L2 in an acute angle, as follows:

- tan ϕ = tan (α2 - α1) = tan α2-tan α11+tan α1 tan α2
- Recalling that the tangent of the acute angle of inclination equals the slope of the line, then we have:
- tan α1 = m1 which stands for the slope of L1 of the acute angle
- tan α2 = m2 which stands for slope of L2 of the acute angle
- such that m2 > m1
- Therefore, substituting the above expression in tangent formula we have: tan ϕ = m2 - m11+ m1 m2

Example: Referring to figure A, find the acute angle that is between the two lines with m, = 12 and m2 = 2 for their slopes

*Solution*:- tan ϕ =2- 12 ÷ 1 + (12)(2) = 34 =.75
- Such that ϕ = arctan (.75) = 360 52’
- If one of the lines in this acute angle was parallel to the Y-axis, then its slope would be infinite and it will render the slope formula for tan + valueless as a result of an infinite value in both the numerator and denominator of the fraction m2 - m11+ m1 m2 produces an intermediate form of an acute angle 1 + m,m2
- However, if only one of the lines of this acute angle is known to be parallel to the Y-axis, the tangent of + can be expressed in another way. Assume that L2 of fig A was parallel to Y-axis. Then we would have something like this; α2 = 900
- But if L1 has a
*positive*slope, then the acute angle that lies between L1 and L2 would be found by; ϕ = 900 – α1 then tan ϕ = cot α1 = 1m1. If L1 of the acute angle has a*negative*slope, then ϕ = 900 – α1 = -(900 - α1) while tan ϕ = - cot α1 = - 1m1