calcule a distancia entre os pontos B e c indicados na figura abaixo.
observação. este numero escrito a caneta não faz parte da imagem não. foi outra pessoa que escreveu isto.
fico muito grato pela atenção e a resposta.
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Bom dia Jose!
Solução!
Vamos usar algumas relações trigonométricas básicas.
![Sen(60\º)= \dfrac{ \sqrt{3} }{2} \\\\\\\\ \dfrac{\overline{DA}}{\overline{CA}}= Cos(60\º)\\\\\\\ \dfrac{15}{ \overline{CA}}= \dfrac{ \sqrt{3} }{2}\\\\\\ \overline{CA}. \sqrt{3} =30\\\\\
\overline{CA}. =\dfrac{30}{\sqrt{3} }\\\\\\
\overline{CA}. =\dfrac{30 \sqrt{3} }{3 }\\\\\\
\overline{CA}= 10 \sqrt{3}](https://tex.z-dn.net/?f=Sen%2860%5C%C2%BA%29%3D+%5Cdfrac%7B+%5Csqrt%7B3%7D+%7D%7B2%7D+%5C%5C%5C%5C%5C%5C%5C%5C+%5Cdfrac%7B%5Coverline%7BDA%7D%7D%7B%5Coverline%7BCA%7D%7D%3D+Cos%2860%5C%C2%BA%29%5C%5C%5C%5C%5C%5C%5C+%5Cdfrac%7B15%7D%7B+%5Coverline%7BCA%7D%7D%3D+%5Cdfrac%7B+%5Csqrt%7B3%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C+%5Coverline%7BCA%7D.+%5Csqrt%7B3%7D+%3D30%5C%5C%5C%5C%5C%0A%5Coverline%7BCA%7D.++%3D%5Cdfrac%7B30%7D%7B%5Csqrt%7B3%7D+%7D%5C%5C%5C%5C%5C%5C%0A%5Coverline%7BCA%7D.++%3D%5Cdfrac%7B30+%5Csqrt%7B3%7D+%7D%7B3+%7D%5C%5C%5C%5C%5C%5C%0A+%0A+%5Coverline%7BCA%7D%3D+10+%5Csqrt%7B3%7D+ "Sen(60\º)= \dfrac{ \sqrt{3} }{2} \\\\\\\\ \dfrac{\overline{DA}}{\overline{CA}}= Cos(60\º)\\\\\\\ \dfrac{15}{ \overline{CA}}= \dfrac{ \sqrt{3} }{2}\\\\\\ \overline{CA}. \sqrt{3} =30\\\\\
\overline{CA}. =\dfrac{30}{\sqrt{3} }\\\\\\
\overline{CA}. =\dfrac{30 \sqrt{3} }{3 }\\\\\\
\overline{CA}= 10 \sqrt{3}")
![\triangle ABC\\\\\ \overline{BC}\\\\\ \overline{AC}=10 \sqrt{3} \\\\\\\ Cos(30\º)= \dfrac{ \sqrt{3} }{2} \\\\\\ \dfrac{\overline{AC}}{\overline{BC}}= Cos(30\º)\\\\\\\ \dfrac{10 \sqrt{3} }{\overline{BC}}=\dfrac{ \sqrt{3} }{2}\\\\\\\ \overline{BC}. \sqrt{3} =20. \sqrt{3} \\\\\\\ \overline{BC}= \dfrac{20 \sqrt{3} }{ \sqrt{3} }\\\\\
\overline{BC}=20](https://tex.z-dn.net/?f=%5Ctriangle+ABC%5C%5C%5C%5C%5C+%5Coverline%7BBC%7D%5C%5C%5C%5C%5C+%5Coverline%7BAC%7D%3D10+%5Csqrt%7B3%7D+%5C%5C%5C%5C%5C%5C%5C+Cos%2830%5C%C2%BA%29%3D+%5Cdfrac%7B+%5Csqrt%7B3%7D+%7D%7B2%7D+%5C%5C%5C%5C%5C%5C+%5Cdfrac%7B%5Coverline%7BAC%7D%7D%7B%5Coverline%7BBC%7D%7D%3D+Cos%2830%5C%C2%BA%29%5C%5C%5C%5C%5C%5C%5C+%5Cdfrac%7B10+%5Csqrt%7B3%7D+%7D%7B%5Coverline%7BBC%7D%7D%3D%5Cdfrac%7B+%5Csqrt%7B3%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C+%5Coverline%7BBC%7D.+%5Csqrt%7B3%7D+%3D20.+%5Csqrt%7B3%7D+%5C%5C%5C%5C%5C%5C%5C+%5Coverline%7BBC%7D%3D+%5Cdfrac%7B20+%5Csqrt%7B3%7D+%7D%7B+%5Csqrt%7B3%7D+%7D%5C%5C%5C%5C%5C%0A+%5Coverline%7BBC%7D%3D20 "\triangle ABC\\\\\ \overline{BC}\\\\\ \overline{AC}=10 \sqrt{3} \\\\\\\ Cos(30\º)= \dfrac{ \sqrt{3} }{2} \\\\\\ \dfrac{\overline{AC}}{\overline{BC}}= Cos(30\º)\\\\\\\ \dfrac{10 \sqrt{3} }{\overline{BC}}=\dfrac{ \sqrt{3} }{2}\\\\\\\ \overline{BC}. \sqrt{3} =20. \sqrt{3} \\\\\\\ \overline{BC}= \dfrac{20 \sqrt{3} }{ \sqrt{3} }\\\\\
\overline{BC}=20")
Bom dia!
Bons estudos!
Solução!
Vamos usar algumas relações trigonométricas básicas.
Bom dia!
Bons estudos!
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