A vertical tower of height 60 m...?

Question

A vertical tower of height 60 m is surmounted by a radio mast of height 25 m. Find the greatest angle that the mast subtends at a point in the horizontal plane through the foot of the tower.

I would like to know from which angles do I obtain the equation of the angle, which will be optimized to find the greatest one.

Answer 1

I don't think Halac's answer is correct. If you start with x = 0, theta = 0. As x increases, theta increases, but reaches a maxim of about 10 degrees, at x = 65, and then starts diminishing.

Oregfiu'a answer is correct. He uses the relationship

tan(a-b) =(tan a - tan b)/1 + tan a* tan b

to derive the tan of the subtended angle.

Then using the arctan, the derivative can be taken with respect to x to find the inflection point. The derivative is messy, but it can be computed. I got exactly the same result as he did.

Actually, there is a second solution to the derivative, that being x = 0. At this point, the angle is zero, and presumably begins increasing as you extend the x in the opposite direction. Interesting.

Answer 2

I'm assuming theta is angle between top of tower and top of antenna.
Let a = angle from bottom of tower to top of tower,
so a = arctan(25/x)
at a distance x from the tower/antenna (on the ground)

tan(a+theta) = (60+25)/x
arctan(25/x) + theta = arctan(85/x)
theta(x) = arctan(85/x) - arctan(25/x)

theta'(x) = -(85/x^2)/[1+(85/x)^2] + (25/x^2)/[1+(25/x)^2]
= 25/[x^2 + 25^2] - 85/[x^2 + 85^2]

theta'(x) = 0, so 25*(x^2 + 85^2) = 85*(x^2 + 25^2)
25 * 85^2 - 85 * 25^2 = 60*x^2
85*25*(85 - 25) = 60*x^2
85*25 = x^2

theta = arctan(sqrt(85)/sqrt(25)) - arctan(sqrt(25)/sqrt(85))
= 33.06deg

Answer 3

x – distance of the point from the axis of the tower
ω - angle that the mast subtends at this point

Relation between x and ω is

tan ω = (25 x) / (x² + 5100)

The greatest angle is ω = 9.93°.
The corresponding distance is x = 71.4 meters.

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To halac:

I think a = arctan(60/x)

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