Answer:
Perimeter is 75.7128129211 units
Step-by-step explanation:
Given ΔАВС, m∠ACB = 90°, CD ⊥ AB and m∠ACD = 30, AD = 8 cm
we have to find the perimeter of ΔABC
In triangle ADC,
[tex]\sin 30^{\circ}=\frac{AD}{AC}=\frac{8}{AC}[/tex]
[tex]\frac{1}{2}=\frac{8}{AC}[/tex] ⇒ [tex]AC=16units[/tex]
and also, [tex]\tan 30^{\circ}=\frac{AD}{CD}=\frac{8}{CD}[/tex]
[tex]\frac{1}{\sqrt{3}}=\frac{8}{CD}[/tex] ⇒ [tex]CD=8\sqrt{3}units[/tex]
Now, in triangle BDC,
∠BDC + ∠ADC = 180°
∠BDC = 180°- 90° = 90°
and also ∠DCB=∠ACB - ∠ACD = 90° - 30° = 60°
[tex]\tan 60^{\circ}=\frac{DB}{CD}=\frac{DB}{8\sqrt{3} }[/tex]
DB=[tex]{\sqrt{3}}\times{8}{\sqrt{3}}[/tex] ⇒ [tex]DB=24units[/tex]
and also [tex]\sin 60^{\circ}=\frac{DB}{BC}=\frac{24}{BC}[/tex]
[tex]\frac{\sqrt{3}}{2}=\frac{24}{BC}[/tex] ⇒ [tex]BC=\frac{48}{\sqrt{3}} units[/tex]
Hence, Perimeter = AC+AD+DB+BC
= 16+8+24+[tex]\frac{48}{\sqrt{3} }[/tex]
= 75.7128129211 units
