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chapter2.3r
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à 2.3ïSurveying Problems
ä Please solve the following surveying problems.
âïTo help determine the cost of flood insurance, a surveyor is
asked to find how high a house is above sea level.ïHe measures from a
point at mean high tide up the slope to the base of the house and gets
a distance of 255 feet.ïThe angle of elevation to the base of the house
is 6.5°.ïFind the elevation above sea level.ïThe elevation above sea
level is 28.87 feet.ïPlease see details for an explanation.
éSêêëWe can use right triangle trigonometry to
êêêïfind the elevation above sea level.
êêêêè sin 6.5°è=èh/255 ft
êêêêêè hè≈è28.87 ft
êêêïThus, the base of the house is 28.87 feet above
@fig2301.bmp,15,25
êêêïsea level.
1ïTo help plan for the construction of a road, a surveyor is
asked to find the number of feet a road will have to ascend in order to
reach the top of a hill.ïHe measures a distance of 1700 feet from the
base to the top of the hill, and an angle of elevation of 3.5°.ïFind
the height of the hill.
ïA)ï103.78 ftëB)ï42.38 ftëC)ï36.22 ftëD)ïå of ç
üêêë The amount of rise can be determined by right
êêêïtriangle trigonometry.
êêêêê sin 3.5°è=èh/1700 ft
êêêêêê hè≈è103.78 ft
êêêïThus, the road will have to rise 103.78 feet.
@fig2302.bmp,100,600
Ç A
2ïA small bridge is to be built across a river from point A to
point B.ïA surveyor measures from point A a distance of 100 feet along
the river bank to point C.ïHe then uses his transit to measure angle A
to be 96.5° and angle C to be 46.8°.ïFind the distance across the river
from point A to point B.
ïA)ï160 ftê B)ï121.98 ftè C)ï110.8 ftëD)ïå of ç
üêêë We can use the Law of Sines when two angles
êêêïand a side are known.ïFirst find angle B to be
êêêï36.7°.ïThen find side "c".
êêêêê 100 ftêë c
#êêêêë ────────è =è ─────────
êêêêë sin 36.7°ê sin 46.8°
@fig2303.bmp,100,600
êêêêêëcè≈è121.98 ft
Ç B
3ïA surveyor needs to measure a distance on a boundry line from
point A to point B, but can't walk directly through the swampy area.
Instead, he sets his transit on dry land at point C and measures angle C
to be 33.7.ïHe also measures CA to be 186.4 feet and BC to be 223.7
feet.ïFind the distance from point A to point B.
ïA)ï229.7 ftêB)ï124.12 ftè C)ï136.8 ftëD)ïå of ç
üêêë The Law of Cosines can be used when two sides
êêêïand the included angle are known.
#êêêïcì = (186.4)ì+(223.7)ì-2∙(186.4)∙(223.7)∙cos33.7°
@fig2304.bmp,100,600
êêêêêè cï≈ï124.12 ft
êêêïThus, the distance across the swamp is 124.12 ft.
Ç B
4ïTo determine the length of a water supply tunnel that must be
dug through a large hill, a surveyor locates a vantage point where he
can see the beginning of the tunnel (point A) and the finish (point B).
He measures the angle at his position, point C, to be 116.8°.ïAlso, CA
is 348 feet, and CB is 526 feet.ïFind the distance AB.
ïA)ï486.1 ftêB)ï583.62 ftè C)ï750.23 ftè D)ïå of ç
üêêë The Law of Cosines can be used when two sides
êêêïand the included angle are known.
#êêêïcì = (348)ì + (526)ì - 2∙(348)∙(526)∙cos 116.8°
@fig2305.bmp,100,600
êêêêêè cï≈ï750.23 ft
êêêïThus, the length of the tunnel will be 750.23 ft.
Ç C
5ïTo determine the height of the Wright Brothers' monument, a
surveyor measures 200 feet down the hill from the base of the monument.
The angle of elevation from that point to the base of the monument is
43.5° and to the top of the monument is 62.8°.ïFind the height of the
monument.
ïA)ï144.62 ftë B)ï123.8 ftëC)ï107.6 ftëD)ïå of ç
üêêë We can use right triangle trigonometry to
êêêïsolve this problem.ïFirst find AC and CD in tri-
êêêïangle ACD.
êêêïcos 43.5° = AC/200êsin 43.5° = CD/200
êêêëAC ≈ 145.075 ftêèCD ≈ 137.67 ft
êêêïThen find BC using triangle ABC.
@fig2306.bmp,100,600
êêêêëtan 62.8°è=èBC/145.075 ft.
êêêêêëBCè≈è282.285 ft
Thus, the monument is (282.285 - 137.67)ï=ï144.62 feet high.
Ç A
6êêèTo establish the location of a corner stake
êêêïfor a lot at point B, a surveyor must determine
êêêïthe angle at C.
êêêêïA)ï76.12°êêB)ï69.34°
@fig2307.bmp,25,118
êêêêïC)ï83.88°êêD)ïå of ç
üïWe know one angle and two sides with one side opposite the given
angle.ïThe Law of Sines can be used in this case.ïAlso, since side "a"
is greater than side b, there is just one solution.ïFirst find angle B.
êêê250 ft/sin Bè=è283 ft/sin 52°
êêêêsin Bè≈è.69612
êêêêè Bè≈è44.12°
Then find angle C.ïangle Cï=è180° - (52° + 44.12°)ï=ï83.88°.
Ç C
7êêèA surveyor is asked to find the area of lot
#êêêïABCD with given dimensions. (1 acre = 43,560 ftì)
êêêêïA)ï1.21 acresêèB)ï1.45 acres
@fig2308.bmp,25,118
êêêêïC)ï1.07 acresêèD)ïå of ç
üïYou can find the area of lot ABCD by finding the area of the
two triangles and adding them together.
#ë Area of triangle ABC = 1/2∙290∙190∙sin 110°ï≈è25,888.53 ftì
#ë Area of triangle ACD = 1/2∙250∙230∙sin 111.68° ≈ 26,716.27 ftì
#êêêêêêêë ─────────────
#ë Total area of lot ABCD in square feetêë52,604.8 ftì
#Thus, the area in acres is 52,604.8 ftì/43,560 ftìï=ï1.21 acres.
Ç A