Pythagoras
and Trigonometry Revision Questions
1.
A. A ship sails 20 km due north and then 35 km due east. How far is
it from its starting point?
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answer...
Use Pythagoras' Theorem:
a^{2} + b^{2} = c^{2}
20^{2} + 35^{2} = c^{2}
400 + 1225 = 1625 = c^{2}
√ 1625 = c = 40.31 km
Note that the question simply asks how far it is from its starting
point, so no bearing is required, just the distance. Don't forget the units! (In this case "km".)
B. At what bearing will the ship need to travel to get back to its
starting point (to the nearest degree)?
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answer...
The ship has traveled north and east, so it will have to go south
and west to get back, so we know the bearing will be somewhere in
the south west quadrant, from 180° to 270°. (In other words,
the answer is a reflex angle.) We use trignometry to find the interior
angle of the triangle, then subtract that from 270°:
tan Ø = ^{opposite} / _{adjacent}
tan Ø = ^{distance travelled north}
/ _{distance travelled east}
tan Ø = ^{20} / _{35}
Ø = tan^{1}( ^{20} / _{35}
)
Ø = 30° (0dp)
bearing = 270°  30° = 240°
Remember, if our answer is outside the southwest quadrant we've
made a mistake somewhere. Don't forget the degrees symbol.
2.
An 18 cm long piece of wire is bent into a U shape by bending it through
90° 3 cm from one end, and another 90° 3 cm further along the
wire. What is the distance between the ends?
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answer...
The area enclosed by the wire can be thought of as a square with
an adjoining rightangle triangle, with the line between the ends
of the wire the hypotenuse of the triangle. Use Pythagoras' theorem,
taking off the length of the three sides of the square to find the
length of the triangle:
a^{2} + b^{2} = c^{2}
3^{2} + (18  3  3  3)^{2} = c^{2}
3^{2} + 9^{2} = c^{2}
9 + 81 = 90 = c^{2}
√ 90 = c = 9.49 cm
3.
A girl is flying a kite with a string length of 60 metres. Ignoring
the girl's height, if the string is taut and makes an angle of 71°
to horizontal, how high is the kite?
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answer...
Use trigonometry:
sin ø = ^{opposite} / _{hypotenuse}
sin ø = ^{height of kite} / _{string
length }
sin (71°) = ^{height of kite} / _{60}
60 * sin (71°) = height of kite
height of kite = 56.73 m
4.
When a particular 2 m ladder is leaned against a wall, its instructions
say it must be placed at a 4:1 angle ratio or steeper to prevent
its feet from slipping outward (ie, for every cm the base is out from
the wall, the top must be at least 4 cm up the wall). What is
the minimum height the ladder will reach up a wall without risking slipping?
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answer...
Use Pythagoras' theorem:
a^{2} + b^{2} = c^{2}
Remember that we square the 4x, not square the x then multiply
by 4.
x^{2} + (4x)^{2} = 2^{2}
x^{2} + 16x^{2} = 17x^{2}
= 4
x^{2} = ^{4} / _{17}
x = √ ( ^{4} / _{17} ) = 0.485
height up wall = 4x = 1.94 m
5.
An archer defending a castle knows he has an accurate/effective range
of 80 m for a humansized target and 160 m for an elephantsized target.
(FWIW Wikipedia
says An archer could hit a person at 165 m (180 yards) "part of
the time" and could always hit an army.) He also knows that when
he is standing on the castle wall, a marker stone 10 m out from the castle
wall makes an angle of about 64° below horizontal, and a marker stone
50 m out makes an angle of about 22° below horizontal.
A. How high is the castle wall?
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answer...
Use trigonometry for each of the marker stones then find an average
of the two values:
tan ø = ^{opposite} / _{adjacent}
tan ø = ^{height of castle wall} /
_{distance of marker stone}
tan (64°) = ^{height of castle wall}
/ _{10}
10 * tan (64°) = height of castle wall = 20.5
m (estimate 1)
tan (22°) = ^{height of castle wall}
/ _{50}
50 * tan (22°) = height of castle wall = 20.2
m (estimate 2)
Average = ( 20.2 + 20.5 ) / 2 = 20.35 m
B. With your answer in whole degrees, what angle below horizontal
will elephants be when the archer can start firing at them? (Consider:
To get an answer in whole degrees should the number be rounded up, down,
or to the nearest whole number?)
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answer...
Use trigonometry:
tan ø = ^{opposite} / _{adjacent}
tan ø = ^{height of castle wall} /
_{distance of elephant}
tan ø = ^{20.35} / _{160}
ø = tan^{1} ( ^{20.35} /
_{160} ) = 7.25°
We should arguably round up to ensure the target is actually within
range  if we rounded down the target would be just outside the maximum
range.
Ø = 8°
C. With your answer in whole degrees, what angle below horizontal
will human targets be when they come within range?
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answer...
Use trigonometry:
tan ø = ^{opposite} / _{adjacent}
tan ø = ^{height of castle wall} /
_{distance of human}
tan ø = ^{20.35} / _{80}
ø = tan^{1} ( ^{20.35} /
_{80} ) = 14.27°
Again, we should arguably round up to ensure the target is within
range.
Ø = 14°
If our answers are more than 22° we've made a mistake, since
that's the angle that objects at 50 m will be at.
