PROGETTO LAUREE SCIENTIFICHE 2013/2014
LICEO P. CALAMANDREI
ALUNNI:
IOV INE RO B ERTA 4 E
M I R A N DA S A LVATO R E 4 E
CIVOLANI CARLO 4G
M ATTEUCCI G IUS EPPE 4 G
MARSIGLIA SEBASTIANO 4F
DAMIANO RICCARDO 4F
VIVO IMMA 4B
MAIONE GIOVANNI 4B
MUNGIELLO GENNARO 4D
CACCAVALE GIOVANNI 4D
CONTE ANDREA 4A
DIAMANTE FRANCESCO 4C
GUADAGNO FEDERICA 4BV
BORRELLI MARTINA 4BV
PAPA DARIO 4AV
CERIELLO GIANLUCA 4AV
Index
1. Problem
2. Geogebra construction
3. Mathematical background
4. Resolution
“Flying off at the tangent”
«It is known that during migration a flock of birds is flying at 260 meters
above the ground. An ornithologist is watching those birds while they are
getting further from her in a straight line, with an elevation angle of 30°. If a
minute later that angle has reduced to 20°, at which speed are the birds
travelling?»
«Si sa che certi uccelli, durante la migrazione, volano ad una
altezza media di 260metri dal suolo. Un’ornitologa osserva
uno stormo di questi volatili, mentre si allontana da lei in
linea retta, con un angolo di elevazione di 30° . Se un
minuto più tardi tale angolo si è ridotto a 20°, con che
velocità si stanno spostando gli uccelli?»
Ornithologist and birds
Mathematical background #1
A rectangle triangle is a particular triangle in which one
of the angles is rect (90°). The opposite side to the
rect angle is called hypotenuse.
Cathetus
90°
Cathetus
Mathematical background #2
Application of one the theorems of rectangle
triangles: in a rectangle triangle, the measure of a
cathetus is equal to the other cathetus times the
tangent of the opposite angle or the cotangent of the
adjacent angle.
Mathematical background #3
The medium speed of a body is equal to
the ratio between the displacement and
the time taken to cover it.
We consider the birds moving in a uniform
linear motion, to solve the problem in an
easier way.
Resolution pt.1
We have drawn a straight line, which correspondes to the X-axis and a
perpendicular in C, to build a right triangle. We have used the button
to draw it.
Resolution pt.2
Next we have considered the trigonometric circle centered in A. To draw it we
have used the button
and selected the value 1 for the radius.
Resolution pt.3
We have considered the point D, intersection between the circle and the right
triangle with the button
. We have drawn the perpendicular line selecting
point D and the X-axis by the button
. We have called E the intersection.
ABC and ADE triangles are similar because they have both a right angle and an
angle α in common
So the following relations are true:
BC:AC=DE:EA
BC/AC=sin(α)/cos(α)=tan(α)
AC/BC=cos(α)/sin(α)=cotan(α)
To solve the problem we have used the right triangle properties : in fact we know
the measure of a cathetus in both triangles ABC and ADE and their opposite
angles. Using the trigonometric formula we have calculated the measure of the
other two catheti :
AB=CB cotg(30°)=CB tg(60°)
AD=ED cotg(20°)=ED tg(70°)
We have calculated the displacement of the birds subtracting the measure of the
side AB from AD:
CE=AD-AB=ED tg(70°)-CB tg(60°)
Finally, we have calculated the speed the birds are travelling at because we know
the displacement and the time covered:
v= CE/t= (ED tg70°-CB tg60°)/60 m/sec= 4,4 m/sec
The speed the birds are travelling at is 4,4 m/s.
THANKS TO
P RO F E S S O R S A LVATO R E C U O M O A N D U L D E R IC O DA R DA N O
FOR THE CHANCE WE WERE GIVEN TO ENRICH OUR
KNOWLEDGE.