Answer:
Perimeter of [tex]\triangle ABC[/tex] is [tex]48+16\sqrt{3}[/tex].
Explanation:
Given: In [tex]\triangle ABC[/tex] , [tex]\angle ACB = 90^{\circ}[/tex] , [tex]\angle ACD= 30^{\circ}[/tex] and AD= 8cm.
In [tex]\triangle ADC[/tex]
Sum of the measure of the angles of triangles is 180 degree.
[tex]\angle CAD+\angle CDA+\angle ACD=180^{\circ}[/tex]
[tex]\angle CAD+90^{\circ}+30^{\circ}=180^{\circ}[/tex] or
[tex]\angle CAD+120^{\circ}=180^{\circ}[/tex]
Simplify:
[tex]\angle CAD=60^{\circ}[/tex]
∵ AD= 8cm and [tex]\angleACD= 30^{\circ}[/tex] to calculate the length of CD we use tangent ratio i.e,
[tex]\tan = \frac{perpendicular}{Adjacent}[/tex]
then;
[tex]\tan 30 = \frac{8}{CD}[/tex] or
[tex]\frac{1}{\sqrt{3}} =\frac{8}{CD}[/tex]
Simplify:
[tex]CD=8\sqrt{3}[/tex] cm.
Also, find the length of AC, we use sine ratio i.e, [tex]\sin=\frac{perpendicular}{Hypotenuse}[/tex].
Then,
[tex]\sin 30 =\frac{AD}{AC}[/tex] or
[tex]\frac{1}{2} =\frac{8}{AC}[/tex]
On simplify:
[tex]AC=2\times 8 =16[/tex]cm.
Now, in triangle ABC;
[tex]\angle BAC= 60^{\circ}[/tex] , [tex]\angle ACB= 90^{\circ}[/tex] then
[tex]\angle ABC= 30^{\circ}[/tex] [Sum of the measure of the angles in the triangle is 180 degree]
To calculate the length of BC;
[tex]\tan A = \frac{BC}{AC}[/tex] [∴[tex]\ tan= \frac{perpendicular}{adjacent}[/tex] ]
therefore,
[tex]\tan 60 = \frac{BC}{16}[/tex]
or
[tex]\sqrt{3}= \frac{BC}{16}[/tex]
Simplify:
[tex]BC=16\sqrt{3}[/tex]cm.
Using Pythagoras theorem in triangle ACB;
Let BD be x cm
AB = 8+x cm , AC = 16 cm and [tex]BC=16\sqrt{3}[/tex] cm
[tex]AB^2=AC^2+BC^2[/tex]
[tex](8+x)^2=(16)^2+(16\sqrt{3} )^2[/tex] or
[tex](8+x)^2=256+768=1024[/tex] or
[tex]8+x=\sqrt{1024} =32[/tex]
Simplify:
x=24 cm
Therefore, the length of AB = 8+x= 8+24=32 cm
Perimeter(P) of triangle ABC is equal to the sum of the sides of the triangle.
⇒ P= AB+BC+AC = [tex]32+16\sqrt{3} +16=48+16\sqrt{3}[/tex] cm
