l

h

xA

B

C

D

If a circle is constructed such that the line drawn back from the try point is a tangent to thecircle, then the point where the circle and the line touch is the position to give the greatestangle at the goal mouth. Any point outside the circle will give an angle smaller, and anypoint inside the circle will give an angle larger, but there is no point that lies inside the circleand lies along this line.Therefore, using the intercept theorem, we say

x2 = h × (h + l)

∴ x =√

h2 + hl

However, using calculus is a bit more tricky.

Let ∠ADB = α and ∠BDC = θ, therefore tan α =h

xand tan (α + θ) =

l + h

x.

Now we follow these mathematical processes.

tan (α + θ) =tan α + tan θ

1 − tan α tan θ

∴l + h

x=

h

x+ tan θ

1 −h tan θ

x

Algebra manipulation leads us to,

tan θ =lx

hl + h2 + x2

We differentiate both sides of this equation with respect to x.

sec2θ ·

dθ

dx=

(

hl + h2 + x2)

l − lx (2x)

(lh + h2 + x2)2

letdθ

dx= 0 ∴ 2lx2 = hl

2 + h2l + lx

2

∴ x2 = hl + h

2

∴ x =√

hl + h2

1