A trapezium ABCD is inscribed into a semicircle of radius l so that the base AD of the trapezium is a diameter and the vertices B & C lie on the circumference. Find the base angle `theta` of the trapezium ABCD which has the greatest perimeter.

Answer

To find the base angle \( \theta \) of the trapezium \( ABCD \) that has the greatest perimeter when inscribed in a semicircle of radius \( l \), we can follow these steps:

### Step 1: Understanding the Geometry
The trapezium \( ABCD \) is inscribed in a semicircle with \( AD \) as the diameter. The points \( B \) and \( C \) lie on the circumference of the semicircle. The angle \( \theta \) is the base angle at points \( A \) and \( D \).

### Step 2: Setting Up the Problem
Let the radius of the semicircle be \( l \). The coordinates of points can be defined as follows:
- \( A(-l, 0) \)
- \( D(l, 0) \)
- The coordinates of points \( B \) and \( C \) can be expressed in terms of \( \theta \).

### Step 3: Finding Coordinates of Points B and C
Using trigonometric identities:
- The height from points \( B \) and \( C \) to the base \( AD \) is \( l \sin \theta \).
- The horizontal distances from the center to points \( B \) and \( C \) can be expressed as \( l \cos \theta \).

Thus, the coordinates of \( B \) and \( C \) are:
- \( B(-l \cos \theta, l \sin \theta) \)
- \( C(l \cos \theta, l \sin \theta) \)

### Step 4: Calculating the Perimeter
The perimeter \( P \) of trapezium \( ABCD \) can be calculated as:
\[ P = AB + BC + CD + DA \]
Where:
- \( AB = \sqrt{((-l \cos \theta) - (-l))^2 + (l \sin \theta - 0)^2} = \sqrt{l^2 (\cos \theta - 1)^2 + l^2 \sin^2 \theta} = l \sqrt{2 - 2\cos \theta} = l \cdot 2\sin\left(\frac{\theta}{2}\right) \)
- \( CD = 2l \cos \theta \)
- \( DA = 2l \)

Thus, the perimeter becomes:
\[ P = 2l \sin\left(\frac{\theta}{2}\right) + 2l \cos \theta + 2l \]

### Step 5: Simplifying the Perimeter Expression
Combining the terms, we can express the perimeter as:
\[ P = 2l \left(1 + \cos \theta + \sin\left(\frac{\theta}{2}\right)\right) \]

### Step 6: Finding the Maximum Perimeter
To find the maximum perimeter, we differentiate \( P \) with respect to \( \theta \) and set the derivative to zero:
\[ \frac{dP}{d\theta} = 0 \]

### Step 7: Solving the Derivative
After differentiating and simplifying, we find:
\[ \sin \theta (2 \cos \theta - 1) = 0 \]
This gives us two cases:
1. \( \sin \theta = 0 \) (which gives angles \( 0 \) or \( \pi \))
2. \( 2 \cos \theta - 1 = 0 \) leading to \( \cos \theta = \frac{1}{2} \) which gives \( \theta = \frac{\pi}{3} \) or \( 60^\circ \).

### Step 8: Confirming Maximum
To confirm that this value gives a maximum, we can check the second derivative or evaluate the perimeter at this angle.

### Final Answer
Thus, the base angle \( \theta \) of the trapezium \( ABCD \) which has the greatest perimeter is:
\[ \theta = 60^\circ \quad \text{or} \quad \frac{\pi}{3} \text{ radians} \]