Respuesta :

calculista

Answer:

[tex]AD=9\ cm[/tex]

Step-by-step explanation:

Step 1

In the right triangle BCD

Find  the measure of CD

we know that

m∠DCB = [tex]90\°-60\°=30\°[/tex]

[tex]cos(30\°)=\frac{CD}{BC}[/tex]

[tex]cos(30\°)=\frac{\sqrt{3}}{2}[/tex]

so

[tex]\frac{CD}{BC}=\frac{\sqrt{3}}{2}[/tex]

solve for CD

[tex]CD=BC\frac{\sqrt{3}}{2}[/tex]

we have

[tex]BC=6\ cm[/tex]

substitute

[tex]CD=(6)\frac{\sqrt{3}}{2}[/tex]

[tex]CD=3\sqrt{3}\ cm[/tex]

Step 2

In the right triangle ACD

Find  the measure of AD

we know that

[tex]tan(60\°)=\frac{AD}{CD}[/tex]

[tex]tan(60\°)=\sqrt{3}[/tex]

so

[tex]\frac{AD}{CD}=\sqrt{3}[/tex]

solve for AD

[tex]AD=(CD)\sqrt{3}[/tex]

we have

[tex]CD=3\sqrt{3}\ cm[/tex]

substitute

[tex]AD=(3\sqrt{3})\sqrt{3}=9\ cm[/tex]

[tex]AD=9\ cm[/tex]

The measurement of the length of AD is 9cm.

What is the property of a right-angle triangle?

There will always be one angle which is 90 degrees. The side that is opposite 90 degrees is known as the hypotenuse.

In the right triangle BCD the measurement of angle m∠DCB is;

[tex]\rm m\angle DCB =90-60=30\\\\Cos \theta=\dfrac{CD}{BC}\\\\Cos30=\dfrac{CD}{BC}\\\\Cos30=\dfrac{\sqrt{3} }{2}[/tex]

On comparing the relations the value of BC is;

[tex]\rm \dfrac{CD}{BC}=\dfrac{\sqrt{3} }{2}\\\\BC=6\\\\ \dfrac{CD}{6}=\dfrac{\sqrt{3} }{2}\\\\ CD= \dfrac{6\sqrt{3} }{2}\\ \\CD = 3\sqrt{3}[/tex]

In the right triangle ACD the measure of the length of AD is;

[tex]\rm Tan60=\dfrac{AD}{CD}\\\\\sqrt{3}=\dfrac{AD}{CD}\\\\AD = \sqrt{3} \times CD\\\\AD = \sqrt{3} \times CD\\\\CD = 3\sqrt{3} \\\\AD = \sqrt{3} \times3\sqrt{3} \\\\ AD = 3\times 3\\\\ AD=9 \\\\[/tex]

Hence, the measurement of the length of AD is 9cm.

To know more about the right-angle triangle click the link given below.

https://brainly.com/question/8517005

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