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)