In the diagram below, we have \(AB = 24\) and \(\angle ADB =90^\circ\). If \( \sin A = \frac23\) and \( \sin C = \frac13\), then what is \(DC\)?
+36916 sin A = 2/3 = BD/24 meaning BD = 16
sin C = 1/3 = BD / BC = 16/BC meaning BC = 48
48^2 - 16^2 = DC ^2 ( Pythag Theorem)
DC = 45.25 units
+208 Hey there!
Since \(sin\) is opposite over hypotenuse, we can use thid to find \(BD\):
\(AB=hyp=24\), \(sin(A)=2/3\), using this ratio we know that \(BD=16\)
We can also use the same stategy to find \(BC\)
\(BD=opp=16\). \(sin(C)=1/3\), using this ration we know that \(BC=48\)
Finally we need to manipulate Pythagoras to give us \(DC\)
\(DC=\sqrt{(BC)^2-(BD)^2}=\sqrt{48^2-16^2}\approx45.3\)
Hope this helped :)
