Question

¿como se resuelve este problema de matematicas?

muchas gracias

Answer 1

$Hola \\ \\ De~la~figura~resolvemos~com~respecto~al~\acute{a}ngulo~de~30\° asi': \\ \\ cos30\°= \frac{SR}{SQ} ~~~---\ \textgreater \ valor~aproximado~de~[cos30\°=0,87~] \\ \\ \underbrace{cos30\°}_{0,87}= \frac{SR}{SQ}~~--\ \textgreater \ y~valor~de~[SQ=250m], reemplazando \\ \\ 0,87= \frac{SR}{250} ~~---\ \textgreater \ despejamos~(SR), nos~queda \\ \\ \boxed{SR=217,5m} \\ \\ ===================================== \\ Continuamos ~com~respecto~al~\acute{a}ngulo~de~30\°, de ~la~figura~calculemos \\ ''QR'', veamos:$

$sen30\°= \frac{QR}{SQ}~~--\ \textgreater \ valor~de~[sen30\°=0,5]~y~[SQ=250~] \\ \\ 0,5= \frac{QR}{250} ~~---\ \textgreater \ despejemos~''QR'' \\ \\ \boxed{ QR=125m } \\ \\ =================================== \\ Ahora~calculemos~com~respecto~al~\acute{a}ngulo~total ~de~''S''~que~es~ \\ 30\°+10\°=40\° \\ De~la~figura~para~calcular~''PQ'', calcularemos~com~respecto \\ al~\acute{a}ngulo~de~40\°, veamos: \\ \\ tg40\°= \frac{PR}{SR} ~~--\ \textgreater \ valor~aproximado~de~[tg40\°=0,84~]$

$0,84= \frac{PR}{SR} ~~--\ \textgreater \ Se~sabe~que~[PR=PQ+QR]~y~[SR=217,5] \\ \\ 0,84= \frac{PQ+QR}{217,5} \\ \\ 0,84(217,5)=PQ+QR~~--\ \textgreater \ se~sabe~tambien~que~[QR=125] \\ \\ 182,7=PQ+125~~--\ \textgreater \ despejado~''PQ'', tenemos \\ \\ PQ=182,7-125 \\ \\ \boxed{\boxed{PQ=57,7m} }$

$======================================= \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Espero~te~sirva~saludos!!$