In a regular pentagon ABCDE, find each angle of ∆ ACD.
Question:
In a regular pentagon ABCDE, find each angle of ∆ ACD.
Solution:

Sum of all interior angles of the pentagon
= (n – 2) × 180°, where n is the number of sides of the pentagon
= (5 – 2) × 180°
= 3 × 180°
= 540°
Therefore, measure of each interior angle = 540°/5 = 108°
Now, in ∆ AED, ∠ AED = 108°
∴ ∠ ADE = (180° – 108°)/2
= 72°/2
= 36°
Similarly, in ∆ ABC, ∠ ABC = 108°
∴ ∠ ACB = (180° – 108°)/2
= 72°/2
= 36°
Therefore, in ∆ ACD, ∠ ADC = 108° – ∠ ADE = 108° – 36° = 72° (Ans)
And, ∠ ACD = 108° – ∠ ACB = 108° – 36° = 72° (Ans)
And, ∠ CAD = 180° – (∠ ADE + ∠ ACB) [Angle sum property of a triangle]
= 180° – (72° + 72°)
= 36° (Ans)