Geometry Problem Solving Question [closed]

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.

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, SaadMar 15 '18 at 0:22

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• Along with question you should also show your effort. – kayush Mar 14 '18 at 17:32
• Now slowly find the lwngths of $BC,BM, BM,$ etc. where $M$ is foot of perpendicular from $B$ to $AC$ – Narasimham Mar 14 '18 at 18:05

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.

• Do i look at the scale factor – user123456 Mar 14 '18 at 17:38
• @user123456 i have added more hints, do check those and try again. If that helps then accept the answer. Thanks – kayush Mar 14 '18 at 17:42
• um this is from a non calculator exam – user123456 Mar 14 '18 at 17:47
• 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 – kayush Mar 14 '18 at 17:49
• i haven't learned about the sin c method yet in school – user123456 Mar 14 '18 at 17:50

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 $DC$: $$DC^2=EC^2-4^2$$ $$DC=\sqrt{\biggl(\frac{40}6\biggr)^2-4^2}=\sqrt{\frac{256}9}=\frac{16}3$$ $$\therefore AD=8-DC=8-\frac{16}3=\frac83$$

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.

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