|
Opgave
In $\Delta ABC$ geldt:
-
$\angle A=30^o$ en $\angle BDC=40^o$
-
$AD=5$
Bereken $BC$ op 1 decimaal nauwkeurig.
|
Uitwerking
$
\begin{array}{l}
\tan (30^\circ ) = \frac{{BC}}{{DB + 5}} \Rightarrow BC = (DB + 5) \cdot \tan (30^\circ ) \\
\tan (40^\circ ) = \frac{{BC}}{{DB}} \Rightarrow BC = DB \cdot \tan (40^\circ ) \\
DB \cdot \tan (40^\circ ) = (DB + 5) \cdot \tan (30^\circ ) \\
DB \cdot \tan (40^\circ ) = DB \cdot \tan (30^\circ ) + 5 \cdot \tan (30^\circ ) \\
DB \cdot \tan (40^\circ ) - DB \cdot \tan (30^\circ ) = 5 \cdot \tan (30^\circ ) \\
DB(\tan (40^\circ ) - \tan (30^\circ )) = 5 \cdot \tan (30^\circ ) \\
DB = \frac{{5 \cdot \tan (30^\circ )}}{{\tan (40^\circ ) - \tan (30^\circ )}} \approx 11,0 \\
\end{array}
$
|
