Question:
The angle of a quadrilateral are in AP whose common difference is 10°. Find the angles.
Solution:
Let the required angles be (a - 15)o, (a - 5)o, (a + 5)o and (a + 15)o, as the common difference is 10 (given).
Then (a - 15)o + (a - 5)o + (a + 5)o + (a + 15)o = 360o
⇒ 4a = 360
⇒ a = 90
Hence, the required angles of a quadrilateral are $(90-15)^{\circ},(90-5)^{\circ},(90+5)^{\circ}$ and $(90+15)^{\circ} ;$ or $75^{\circ}, 85^{\circ}, 95^{\circ}$ and $105^{\circ}$.