Specialist Maths - Trigonometry Applications

Question

A BMX track is surrounded by a circular fence of radius 120 metres. Find the percentage of fence visible to a person standing outside the fence, 80 metres from the nearest point of the fence.

It has something to do with peripheral vision and the answer is 29.5%

Answer 1

If you are standing outside the circle the picture would look something like this:

Since you are 80 m outside the circle of radius 120 m, you are 200 m from the center of the circle. The line tangent to the circle marks the maximum arc you can see (from A to B). The line from the center of the circle to point B not only intersects at the tangent but the tangent line forms a right angle with the radius. So, because the triangle formed is a right triangle, you can say the sin(θ)=120/200 or θ = sin^-1(120/200) = 36.87°

The entire angle (∠ADB) is double angle θ because the triangle is the same on the opposite side (point A is the mirror image of point B). So the ∠ADB = 2(36.87) = 73.74°.

There is a secant theorem that states the m∠ADB = 1/2[m(arc ACB) - m(arc AB)] so:

73.74 = 1/2[m(arc ACB) - m(arc AB)] or

2(73.74) = m(arc ACB) - m(arc AB)

However m(arc ACB) + m(arc AB) = 360 so we can flip that around a bit to write that

m(arc ACB) = 360 - m(arc AB). Substituting into the equation above we can say:

2(73.74) = [360 - m(arc AB)] - m(arc AB) and then simplifying:

147.48 = 360 - 2(m(arc AB))

147.48 - 360 = -2(m(arc AB))

-212.562 = -2(m(arc AB))

m(arc AB) = 106.26°

So you can see an arc of 106.26°

Now, since the entire circle is 360°, the percentage you can see is 106.26/360 = 0.295 or 29.5%