+0  
 
0
48
3
avatar

In the diagram below, we have \(AB = 24\) and \(\angle ADB =90^\circ\). If \( \sin A = \frac23\) and \( \sin C = \frac13\), then what is \(DC\)?

 

 Mar 13, 2021
 #1
avatar+31281 
+2

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

 Mar 13, 2021
 #2
avatar
+1

Thank you! Also if anyone needs the exact form instead of the rounded form its \(32\sqrt{2}\)

Guest Mar 13, 2021
 #3
avatar+207 
+3

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 :)

 Mar 13, 2021

21 Online Users

avatar
avatar
avatar