Geometry Problem Solving Question

Question

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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Answer 1

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.

Answer 2

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.

Answer 3

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