We know that,

1 ≤ | sin x | + | cos x | ≤

∴ y = [| sin x | + | cos x |] = 1

Let P and Q be the points of intersection of given curves.

Clearly the given curves meet at points where y = 1

so, we get

x 2 + 1 = 5

x = ±2

Now, P (2, 1) and Q (–2, 1))

Now, x 2 + y 2 = 5

Differentiating the above equation w.r. t. x, we get

2x + 2y = 0

⇒ = –

= – 2 and = 2

Clearly the slope of line y = 1 is zero and the slope of the tangents at P and Q are (–2) and (2) respectively.

Thus, the angle of intersection is tan –1 (2).