Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**austin81****Member**- Registered: 2005-03-21
- Posts: 39

A cylindrical tower 7.5m high is surmounted by a dome 6m in diameter, on the top of which is a spike. A boy whose eye is 1.5m from the ground can just see the tip of the spike over the curve of the dome when his line of sight is at an elevation of 40°.Calculate the length of the spike and the distance the boy is from the nearest point of the tower.

Please I really need this help

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,569

Egads and Gazooks! Give me a coupla days!

This is how I would start:

Forget all the 3 dimensional stuff, it is really a 2d puzzle, so just write it down on a sheet of paper:

| (spike)

.-' (circle 6m diameter)

.-' ^

eye------------ 7.5m

1.5m

---+-----------------------------------

Draw a circle 6m in diameter with its centre 7.5m up. Draw a line (eye height) 1.5m up. Now draw a line at 40° that just grazes the circle.

You could now measure off the paper how far away the boy is standing (by where the 40° and the "eye-line" intersect), and the length of the spike (the distance from the top of the circle upwards to the 40° line), but we should really do all the math!

The puzzler has made it easy for us, I think, by having the circle's diameter 6m, which is exactly the same as the difference in heights (7.5m-1.5m=6m), or else it is just coincidence

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,569

Because the circle sits at eye-level, and also has the 40° line grazing its upper surface, then a line at 20° (half of 40°) would go thru the circles center.

This makes a triangle ABC with A being the eye, B being the base of the circle and C being the center of the circle. We know the circle has an angle at A of 20°, and line BC is 3m (half of the 6m diameter).

Solving this triangle: tan 20° = 3m / line length (AB), so AB = 3m / tan 20° = 8.24 m

The question asks "the distance the boy is from the nearest point of the tower.", well we have just figured out that the boy is 8.24m from the centre line of the tower, the tower is 6m in diameter, so he is 8.24 - 3 = 5.24m away from the outer wall

Now all that is left is that spike !

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**JustaGuest****Guest**

Unless I'm mistaken you seem to have missed what the problem is asking.

I could be wrong.

I read the question as the boy's line of sight doesn't touch the top of the dome - it passes somewhere else. That is we are considering a triangle with angle of elevation 40 degrees, meeting a circle at a tangent (probably not at the top).

Thus I would reduce the problem problem to one using similar triangles, by pythagoras, then use a triangle starting at a height 7.5m where the top angle of the triangle is 90 degrees (tangent to a circle).

Finding the height of the spike is merely similar triangles.

Finding the distance from the tower will involve finding the centre of the tower.

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,569

Thank you. I did not mean that his line of sight passed through the very top, I merely said "grazes the upper surface" (ie a tangent)

Yes, it needs triangles to solve. And the spike's height can likewise be solved.

Could you crunch the numbers, and see if you agree with my 8.24m distance to centre line of dome?

Oh, and if you can spare the time you are most welcome to drop by the forum every so often and help with some of the "Help Me!" problems.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**John E. Franklin****Guest**

When I give my answer below as the distance to the column, I am assuming that the dome fits perfectly on the column,

so the column has a radius of three meters.

(3 + spike)cos 40°=3, ∴ spike height = 0.91622 meters.

tan 40°=(3sin50°+6)/((dist_from_column)+(3-3cos50°)), ∴ boy to column is 8.81769 meters.

For a double check, tan40°=rise over run=(7.5-1.5+3+spike_height)/(boy_to_column + column_radius_of_3).