ΔDEF , G ligt op EF, ∠DGF= θ
te bewijzen: EFcotθ=GFcotE - EGcotF
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sensor
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- Lid geworden op: vr 27 jan 2012, 11:42
Re: cot
Klopt dit?
- In \( \Delta DGF \): \( \cot(\theta) = \frac{DG}{GF} \), dus \( DG = GF \cot(\theta) \).
- In \( \Delta DGE \): \( \cot(E) = \frac{DE}{EG} \), dus \( DE = EG \cot(E) \).
- In \( \Delta DGF \): \( \cot(F) = \frac{DF}{FG} \), dus \( DF = GF \cot(F) \).
- In \( \Delta DGF \): \( \cot(\theta) = \frac{DG}{GF} \), dus \( DG = GF \cot(\theta) \).
- In \( \Delta DGE \): \( \cot(E) = \frac{DE}{EG} \), dus \( DE = EG \cot(E) \).
- In \( \Delta DGF \): \( \cot(F) = \frac{DF}{FG} \), dus \( DF = GF \cot(F) \).
