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careva13
user (0)
Pe: 2011-04-13T18:59:46+03:00 In: In:

Geometrie-Trigonometrie

1) Ipoteza: /_\ABC -oarecare ; BC=10 ; tgB=2 ; tgC=3
Concluzie: AB=? ; AC=? ; tgA=?

Nu prea am inteles cum aflu tgA… 🙄

2) Ipoteza: /_\ABC ; BC=8 ; m.unghiului C=60% ; m.unghiului B=75%
Concluzie: AB, AC, Aria /_\ABC

Scuze ca am pus 2 Probleme dar .. am nev neaparat.

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  1. ambri
    veteran (III)

    \rm{Ducem AD\bot{BC}\\Din \Delta ABD \Rightarrow tgB=\frac{AD}{BD}, dar tgB= 2 \Rightarrow \frac{AD}{BD}=2 \Rightarrow \frac{BD}{AD}=\frac{1}{2} (1) \\Din \Delta ACD \Rightarrow tgC=\frac{AD}{CD}, dar tgC= 3 \Rightarrow \frac{AD}{CD}=3 \Rightarrow \frac{CD}{AD}=\frac{1}{3} (2) \\(1), (2) \Rightarrow \frac{BD}{AD}+\frac{CD}{AD}=\frac{1}{2}+\frac{1}{3} \Rightarrow \frac{BD+CD}{AD}=\frac{5}{6}\Rightarrow \frac{BC}{AD}=\frac{5}{6} \Rightarrow \frac{10}{AD}=\frac{5}{6}\Rightarrow AD=12cm (3)\\(1), (3) \Rightarrow BD=6cm\\(2), (3) \Rightarrow CD=4cm.\\Th.Pitagora in \Delta DAB \Rightarrow AB=6\sqrt5 cm.\\Th.Pitagora in \Delta DAC \Rightarrow AC=4\sqrt{10}cm\\\mathcal {A}_{ABC} =\frac{BC\cdot AD}{2}=\frac{10\cdot12}{2}\Rightarrow \mathcal{A}_{ABC}=60cm^2 (4)\\Dar, \mathcal{A}_{ABC}=\frac{AB\cdot AC\cdot sinA}{2}=60\sqrt2\cdot sinA (5)\\(4), (5) \Rightarrow sinA=\frac{\sqrt2}{2}\Rightarrow m(\hat A)=45^0 \Rightarrow tgA=tg45^0=1. \bl }

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