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Sally is a surveyor in the badlands in South Dakota. At point A on the south side of a deep canyon, she sights across to a fire tower on the north side of the canyon at point B. Then she moves a measured 500 feet along the south rim of the canyon to a point C on a line perpendicular to her original sight line AB. Setting up her transit again, she finds that her line of sight, CB, to the tower is 84 degrees. Correct to the nearest whole foot, how far is point C from the tower?

2007-01-28 02:15:04 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

3 answers

Let P be the point vertically below B on the canyon floor (vertical projection of B onto plane

AP and AC are perpendicular =>
ACP is a right triangle with right angle at A.

|AC| = 500
CB (hypotenuse) makes 84 deg with AP, CP
cos 84 = |CP|/|CB|

2007-01-28 02:21:05 · answer #1 · answered by smci 7 · 0 1

You are correct. To the nearest foot the distance from C to A is 4783 feet. You have a right triangle ABC, where AB is from the original setup point (A) to the tower (B), and AC is the 500 foot leg to the sighting point (C). Since the angle is 84 degrees (assuming from the line AC) the equation becomes:
cos(84 degrees) = AC/CB = 500/CB or
CB = 500/cos(84 degrees) = 4783.386

2007-02-01 01:51:26 · answer #2 · answered by Anonymous · 1 0

This is a basic trig problem, I forgot my TI86 but the formula is goes like this; let a = 500 feet

a^2 + (a (tan 84))^2 = c^2 distance = sqr(c^2)

in english : 500^2 + (500 * tan 84)^2 = c^2

the formula in the parenthesis gives you the distance betwen AB and then adding the suqares of both distances gives you the square of the distance CB. This is known as the pythagorean theorem, a^2 + b^2 = c^2.

2007-01-28 02:56:42 · answer #3 · answered by ikeman32 6 · 0 0

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