Question

Question 2

Answer 1

Answer:

$CD=(3-\sqrt{3})\ cm$

Step-by-step explanation:

step 1

Find the length side AB

In the right triangle ABD

$sin(30\°)=\frac{AB}{AD}$

$AB=sin(30\°)(AD)$

we have

$AD=2\sqrt{3}\ cm$

$sin(30\°)=\frac{1}{2}$

substitute

$AB=\frac{1}{2}(2\sqrt{3})$

$AB=\sqrt{3}\ cm$

step 2

Find the length side BD

In the right triangle ABD

$cos(30\°)=\frac{BD}{AD}$

$BD=cos(30\°)(AD)$

we have

$AD=2\sqrt{3}\ cm$

$cos(30\°)=\frac{\sqrt{3}}{2}$

substitute

$BD=\frac{\sqrt{3}}{2}(2\sqrt{3})$

$BD=3\ cm$

step 3

Find the length side BC

In the right triangle ABC

we know that

BC=AB -----> is an 45°-90°-45° triangle

therefore

$BC=\sqrt{3}\ cm$

step 4

Find the length side CD

we know that

$BD=BC+CD\\CD=BD-BC$

substitute the values

$CD=(3-\sqrt{3})\ cm$