Meghan Rees
Whitney Hansley
Alex Kaskel
6.7, #61
To solve problems a and b, Figure 6.90 is needed:
a) D is the distance from the crow’s nest on the ship’s mast, point C, to the horizon, point
H.
The radius, r, forms a 90
°
angle with point H.
We now have
∆
HCO.
Using the
Pythagorean Theorem, a
2
+ b
2
= c
2
, where a and b are the two legs and c is the
hypotenuse, we get the formula:
d
2
+ r
2
= (r+x)
2
r
2
+2rx+x
2
d
2
= 2rx + x
2
d =
2rx+x
2
The distance from the ship, S, to the horizon, H, along the Earth’s surface is represented
by l.
The angle from the center of the Earth,
ϑ
, is equal to the corresponding surface on
the Earth, l.
According to Sohcahtoa,
sin = opposite
.
/. hypotenuse
sin (
ϑ
) = d
.
/. (x+r) = l
The radius is shown (twice) as a line from the center of the Earth, O, to the surface of the
Earth, S and H.
It is represented by r.
Again, with
∆
HCO, we use the Pythagorean
Theorem to get the formula:
r
2
+ d
2
= (r+x)
2
d
2
= 2xr + x
2
d
2
– x
2
= r
2x
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 Fall '08
 BLAKELOCK
 Calculus

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