Consider an obtuse angled triangle A B C in which the difference between the largest and the smallest angle is π 2 and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1 . (There are two questions based on PARAGRAPH   "   1   " , the question given below is one of them) Then the inradius of the triangle A B C is

Step-by-step solution

Let angle A be obtuse angle and let sides be a - d ,   a ,   a + d So, plotting the diagram we get, Now let sides be, a − d , a , a + d Also given angle A − C = π 2 And circum radius R = 1 Now by sine rule we get, a + d sin ⁡ A = a sin ⁡ B = a − d sin ⁡ C = 2 ∵ A = π 2 + C ⇒ sin ⁡ A = sin ⁡ π 2 + C ⇒ sin ⁡ A = cos ⁡ C ⇒ a + d 2 = 1 − sin 2 ⁡ C ⇒ a + d 2 2 = 1 - a - d 2 2 ⇒ 2 a 2 + d 2