Given: ΔАВС, m∠ACB = 90° CD ⊥ AB, m∠ACD = 60° BC = 6 cm Find CD, Area of ΔABC

Answer:CD = 5.196 cmArea = 31.177 sq. cm.Step-by-step explanation:See the attached diagram. Given that ∠ ACB = 90° in Δ ABC. Now, CD ⊥ AB and ∠ CDB = ∠ CDA = 90° Given that ∠ ACD = 60° and BC = 6 cm. We have to find the length of CD and the area of Δ ABC. Now, ∠ CAD = 90° - ∠ ACD = 90° - 60° = 30°  Again, ∠ CBD = 90° - ∠ CAD = 90° - 30° = 60°. Now, from Δ BCD, [tex]\sin 60 = \frac{CD}{BC} = \frac{CD}{6}[/tex] {Since Δ BCD is a right triangle and ∠ CDB = 90°}  ⇒ [tex]CD = 6 \times \sin 60 = 5.196[/tex] cm. (Answer) Now, from Δ ACD, [tex]\sin 30 = \frac{CD}{AC} = \frac{5.169}{AC}[/tex] {Since Δ ACD is a right triangle and ∠ ADC = 90°} ⇒ [tex]AC = \frac{5.196}{\sin 30} = 10.392[/tex] cm So, the area of Δ ABC = [tex]\frac{1}{2} \times BC \times AC = \frac{10.392 \times 6}{2} = 31.177[/tex] sq. cm. (Answer)