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# need help on this question

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48
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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
+31281
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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
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Thank you! Also if anyone needs the exact form instead of the rounded form its $$32\sqrt{2}$$

Guest Mar 13, 2021
#3
+207
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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