Matemáticas, pregunta formulada por Marc3laa, hace 5 meses

Calcular la altura de una arbol sabiendo que desde un punto del terreno se observa su copa bajo un angulo de 30grados y si nos acercamos 10m bajo un angulo de 60grados

Respuestas a la pregunta

Contestado por charajahancc
0

Respuesta:

/|

/ / |

/ / |

/ / |

/ / | h=?

/ / |

/ / |

/ )30° /)60° |

|--10m---|----x------|

\begin{gathered}\mathbb{uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu} \\ De~la~figura~primero~vamos~calcular~el ~valor~de(x)~con~respecto \\ al~\acute{a}ngulo~de~60\°~vea: \\ \\ tg60\°= \frac{h}{x} ~~~--\ \textgreater \ valor~aproximado~de~[tg60\°=1,7] \\ \\ 1,7= \frac{h}{x} ~~--\ \textgreater \ despejando~(x)~tenemos: \\ \\ \boxed{x= \frac{h}{1,7} }~-----\ \textgreater \ (I) \\ \\ \mathbb{uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu} \\ \\ Ahora~calculamos~con~respecto~al ~\acute{a}ngulo~de~30\°~vea:\\ \\ \end{gathered}

uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

De la figura primero vamos calcular el valor de(x) con respecto

al

a

ˊ

ngulo de 60\° vea:

tg60\°=

x

h

−− \textgreater valor aproximado de [tg60\°=1,7]

1,7=

x

h

−− \textgreater despejando (x) tenemos:

x=

1,7

h

−−−−− \textgreater (I)

uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

Ahora calculamos con respecto al

a

ˊ

ngulo de 30\° vea:

\begin{gathered}tg30\°= \frac{h}{10+x} ~~--\ \textgreater \ valor~aproximado~de~[tg30\°=0,6] \\ \\ 0,6= \frac{h}{10+x}~~--\ \textgreater \ despejamos~(h)~tenemos: \\ \\ \boxed{h=0,6(10+x) }~~-------\ \textgreater \ (II) \\ \\ \mathbb{uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu} \\ Ahora~para~calcular~el~valor~de~(h)~reemplazamos~(I)~en~(II)~vea: \\ \\ h=0,6(10+x)~~--\ \textgreater \ sabemos~que~[x= \frac{h}{1,7}] \\ \\ h=0,6(10+ \frac{h}{1,7})~~--\ \textgreater \ operando~queda: \\ \\ h=6+0,4h \\ \\ \end{gathered}

tg30\°=

10+x

h

−− \textgreater valor aproximado de [tg30\°=0,6]

0,6=

10+x

h

−− \textgreater despejamos (h) tenemos:

h=0,6(10+x)

−−−−−−− \textgreater (II)

uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

Ahora para calcular el valor de (h) reemplazamos (I) en (II) vea:

h=0,6(10+x) −− \textgreater sabemos que [x=

1,7

h

]

h=0,6(10+

1,7

h

) −− \textgreater operando queda:

h=6+0,4h

\begin{gathered}h-0,4h=6 \\ \\ 0,6h=6 \\ \\ h= \frac{6}{0,6} \\ \\ \boxed{h=10m}~~--\ \textgreater \ ~altura~del~arbol \\ \\ \mathbb{uuuuuuuuuuuuuuuuuuuuuuuuuuuuuu} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~Espero~te~sirva~saludos!! \\ \\ \end{gathered}

h−0,4h=6

0,6h=6

h=

0,6

6

h=10m

−− \textgreater altura del arbol

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Espero te sirva saludos!!

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