If the lengths of the sides are \(1, \sin x, \cos x\) in a triangle \(A B C\) then the greatest value of the angle in \(\triangle A B C\) is \([0

We are given a triangle ABC with sides 1, sin(x), and cos(x). From the triangle inequality theorem, we know that in any triangle, the sum of the lengths of any two sides must be greater than the length of the third side: 1. \(1 + \sin x > \cos x\) 2. \(1 + \cos x > \sin x\) 3. \(\sin x + \cos x > 1\) Now, from the given range of \(x\), we know that \(\sin x\) and \(\cos x\) are positive, so \(\sin x > 0\) and \(\cos x > 0\).

As the range of \(x\) is given as \(0 < x < \dfrac{\pi}{2}\), we have the following properties of sine and cosine: 1. \(0 < \sin x \le 1\) 2. \(0 < \cos x \le 1\) When analyzing the range of angles in triangle ABC: 1. \(\sin x > 1 - \cos x\) 2. \(\cos x > 1 - \sin x\)

We will now use the sine law to find the greatest angle: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\) Let \(A = x\), \(a = \sin x\): \(\dfrac{\sin x}{\sin x} = \dfrac{1}{\sin B} = \dfrac{\cos x}{\sin C}\) 1. \(\dfrac{1}{\sin B} = \dfrac{\cos x}{\sin C}\) Now, the greatest angle is the one with the least sine value: 2. \(\sin B > \sin C\) From equations (1) and (2): 3. \(\dfrac{1}{\sin B} < \dfrac{1}{\sin C}\) To satisfy this inequality, the value of \(\sin B\) must be as small as possible and \(\sin C\) must be as large as possible: \(\sin B = 1 - \cos x\) \(\sin C = 1 - \sin x\) Since \(0 < \cos x, \sin x \le 1\), the smallest value of \(\sin B\) and the largest value of \(\sin C\) can be obtained when \(\cos x\) and \(\sin x\) are equal, respectively: \(1 - \cos x = \sin x\) \(1 - \sin x = \cos x\) As \(x = \dfrac{\pi}{2}\), the greatest angle of the triangle must be \(\dfrac{\pi}{2}\).

From the previous steps, we found the greatest angle of the triangle to be \(\dfrac{\pi}{2}\). Therefore, the correct answer is: (a) \(\dfrac{\pi}{2}\)