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I am struggling with this question where I don't know how to use my knowledge on Pythagoras theorem and trigonometry to work out the missing length.

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Thank You and help is appreciated

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closed as off-topic by Xander Henderson, GNUSupporter 8964民主女神 地下教會, Strants, Brandon Carter, Saad Mar 15 '18 at 0:22

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  • $\begingroup$ Along with question you should also show your effort. $\endgroup$ – kayush Mar 14 '18 at 17:32
  • $\begingroup$ Now slowly find the lwngths of $BC,BM, BM, $ etc. where $M$ is foot of perpendicular from $B$ to $AC$ $\endgroup$ – Narasimham Mar 14 '18 at 18:05
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Hint: Try applying similarity or equating $\tan c$ in both triangles.

More Hints:

$1$: Find $BC$ using Pythagoras Theorem.

$2$: Then find $\tan c$ for both triangles and equate, you will get $CD$.

With Similarity:
Notice in $\triangle EDC$ and $\triangle ABC$
$\angle DCE = \angle BCA and \angle EDC = \angle ABC = 90$
$\triangle DCE ~ \triangle BCA$ by AA similarity Now, $\cfrac {ED}{DC} = \cfrac{AB}{BC}$ And you can get $BC$ by applying pythagoras theorem.

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  • $\begingroup$ Do i look at the scale factor $\endgroup$ – user123456 Mar 14 '18 at 17:38
  • $\begingroup$ @user123456 i have added more hints, do check those and try again. If that helps then accept the answer. Thanks $\endgroup$ – kayush Mar 14 '18 at 17:42
  • $\begingroup$ um this is from a non calculator exam $\endgroup$ – user123456 Mar 14 '18 at 17:47
  • $\begingroup$ You don't need a calculator, in $\triangle ABC \tan C = \cfrac{AB}{BC}$ and in $\triangle CED \tan C =\cfrac{ED}{CD}$ , just equate these two fractions to get CD $\endgroup$ – kayush Mar 14 '18 at 17:49
  • $\begingroup$ i haven't learned about the sin c method yet in school $\endgroup$ – user123456 Mar 14 '18 at 17:50
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Using the law of sines, you know that: $$\sin C=\frac6{10}$$ And since $C$ is common to $\triangle ABC$ and $\triangle EDC$, then $EC$ can be solved as: $$\sin C=\frac{4}{EC}\Rightarrow EC=\frac4{\sin C}=\frac4{\frac6{10}}=\frac{40}6$$ And using the pythagorean theorem, we can now solve for $DC$: $$DC^2=EC^2-4^2$$ $$DC=\sqrt{\biggl(\frac{40}6\biggr)^2-4^2}=\sqrt{\frac{256}9}=\frac{16}3$$ $$\therefore AD=8-DC=8-\frac{16}3=\frac83$$

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Note that by the pythagorean theorem, $BC=8$.

And $\triangle BCA$ is similar to $\triangle DCE$ due to $AA$ (Angle-Angle) Similarity.

Look at $DE$. It is similar to $AB$.

Hint: One triangle has sides $\dfrac 32$ times greater than the other.

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  • $\begingroup$ um do I find the scale factor $\endgroup$ – user123456 Mar 14 '18 at 17:41
  • $\begingroup$ Yes, one triangle has sides $\dfrac 32$ greater than the other. @user123456 $\endgroup$ – user535339 Mar 14 '18 at 17:43
  • $\begingroup$ um as BC is 8cm would the scale factor from BC to DC be 2 as 4cm to 8cm is x2 scale factor? $\endgroup$ – user123456 Mar 14 '18 at 17:52
  • $\begingroup$ oh nvm i didn't realise you had to flip the other triangle too $\endgroup$ – user123456 Mar 14 '18 at 17:56
  • $\begingroup$ now do I do 8cm / 3/2 to work out length $\endgroup$ – user123456 Mar 14 '18 at 17:59

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