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PROGETTO LAUREE SCIENTIFICHE 2013/2014 LICEO P. CALAMANDREI ALUNNI: IOVINE ROBERTA 4E MIRANDA SALVATORE 4E CIVOLANI CARLO 4G MATTEUCCI GIUSEPPE 4G 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

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Index 1.Problem 2.Geogebra construction 3.Mathematical background 4.Resolution

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PROBLEM

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“Flying off at the tangent ” «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?» «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?»

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Ornithologist and birds

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GEOGEBRA CONSTRUCTION

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MATHEMATICAL BACKGROUND

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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. Hypotenuse Cathetus 90°

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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.

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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.

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RESOLUTION

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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.

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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.

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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.

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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)

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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.

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THANKS TO PROFESSOR SALVATORE CUOMO AND ULDERICO DARDANO FOR THE CHANCE WE WERE GIVEN TO ENRICH OUR KNOWLEDGE.

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