Question

Solve the problems below. Please answer with completely simplified exact value(s) or expression(s). Given: ΔАВС, m∠ACB = 90 CD ⊥ AB , m∠ACD = 30°,AD = 8 cm. Find: Perimeter of ΔABC

Answer 1

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,

$\sin 30^{\circ}=\frac{AD}{AC}=\frac{8}{AC}$

$\frac{1}{2}=\frac{8}{AC}$ ⇒ $AC=16units$

and also, $\tan 30^{\circ}=\frac{AD}{CD}=\frac{8}{CD}$

$\frac{1}{\sqrt{3}}=\frac{8}{CD}$ ⇒ $CD=8\sqrt{3}units$

Now, in triangle BDC,

∠BDC + ∠ADC = 180°

∠BDC = 180°- 90° = 90°

and also ∠DCB=∠ACB - ∠ACD = 90° - 30° = 60°

$\tan 60^{\circ}=\frac{DB}{CD}=\frac{DB}{8\sqrt{3} }$

DB=${\sqrt{3}}\times{8}{\sqrt{3}}$ ⇒ $DB=24units$

and also $\sin 60^{\circ}=\frac{DB}{BC}=\frac{24}{BC}$

$\frac{\sqrt{3}}{2}=\frac{24}{BC}$ ⇒ $BC=\frac{48}{\sqrt{3}} units$

Hence, Perimeter = AC+AD+DB+BC

                             = 16+8+24+$\frac{48}{\sqrt{3} }$

                            = 75.7128129211 units