Q.An arc is in the form of a parabola with its axis vertical. The arc is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?
Angle Between Two Lines – From Intuition to Precision
When you think of two lines crossing each other, the first thing you notice is how "wide" or "narrow" the opening between them is. That opening is the angle between the lines. If you hold two pens and let them cross, the smaller turn you make to bring one pen onto the other is the angle between them.
But here's the key: two intersecting lines actually make four angles — two acute (sharp) and two obtuse (wide), or all four right angles if they are perpendicular. By convention, when we say "the angle between two lines," we always mean the smaller (acute) angle, which lies between 0∘ and 90∘. If the lines are parallel, the angle is 0∘; if they are perpendicular, it is 90∘.
The Geometry of Slopes
Every non-vertical line in the coordinate plane has a slopem, which tells you how steep it is. The slope is the tangent of the angle the line makes with the positive x-axis. So if a line makes an angle θ with the x-axis, then m=tanθ.
Now imagine two lines with slopes m1 and m2. They make angles θ1 and θ2 with the x-axis. The angle between the lines themselves is simply the difference between these two angles: ∣θ1−θ2∣.
tanϕ=1+m1m2m1−m2
Here ϕ is the acute angle between the two lines. The absolute value ensures we get the smaller angle. The denominator 1+m1m2 comes from the tangent subtraction formula: tan(θ1−θ2)=1+tanθ1tanθ2tanθ1−tanθ2.
Why the Formula Works
Suppose line L1 has slope m1=tanθ1 and line L2 has slope m2=tanθ2. The angle between them is ϕ=∣θ1−θ2∣. Using the tangent subtraction identity:
The absolute value guarantees we take the acute angle. If 1+m1m2=0, the denominator is zero, meaning tanϕ is undefined — that happens when ϕ=90∘, i.e., the lines are perpendicular.
Watch out
If 1+m1m2=0, do not use the formula directly. The lines are perpendicular, so ϕ=90∘. The formula simply tells you the angle is 90∘ by giving an undefined tangent.
Special Cases
Parallel lines: m1=m2. Then numerator is zero, so tanϕ=0, giving ϕ=0∘.
Perpendicular lines: m1m2=−1. Then denominator is zero, so ϕ=90∘.
One vertical line: A vertical line has no defined slope (infinite). If one line is vertical, the angle between it and a line of slope m is 90∘−arctan(m) (or its complement). The formula above does not apply directly; you handle this case separately.
A Quick Example
Find the acute angle between the lines y=2x+3 and y=−3x+1.
Here m1=2, m2=−3.
tanϕ=1+(2)(−3)2−(−3)=1−65=−55=1
So tanϕ=1, which means ϕ=45∘.
Tip
Always check if the denominator is zero first. If it is, the answer is 90∘ and you're done. If not, plug into the formula.
The Big Picture
The angle between two lines is a measure of their relative orientation. The formula tanϕ=1+m1m2m1−m2 is your tool for finding it when you have slopes. It comes directly from the geometry of angles and the tangent subtraction identity — nothing more than that.
The acute angle ϕ between two lines with slopes m1 and m2 is given by tanϕ=1+m1m2m1−m2, with ϕ=90∘ when 1+m1m2=0.
Modelling the arc as a downward-opening parabola with its vertex at the top, the base data (height and width) fixes the parabola's equation, which then gives the width at any distance from the vertex.
✓Final answer
The arc is 5≈2.236 m wide, 2 m from the vertex.
Model the arc as a downward-opening parabola with vertex at the top; use the base data to fix the parabola, then find the width 2 m below the vertex.
Parabola with vertex at origin, axis vertical, opening downward (measuring y as distance dropped from the vertex): x2=4ay.
Place the vertex (top of the arc) at the origin, with y measured as the vertical drop from the vertex and x the horizontal half-width at that drop.
At the base, the drop is y=10 m (the full height of the arc) and the half-width is x=25=2.5 m (base width 5 m).
Substitute (2.5,10) into x2=4ay: (2.5)2=4a(10)⇒6.25=40a⇒a=406.25=325.
Parabola: x2=4(325)y=85y.
At y=2 m from the vertex: x2=85(2)=45⇒x=45=25 m.
Full width at that level =2x=2×25=5≈2.236 m.
Self-check: at y=10, x2=85(10)=6.25⇒x=2.5✓, matching the base half-width.