The prompt invites students to use the formula for the area of a nonrightangled triangle:
During classroom inquiry, students' initial questions and observations have included:
 How do you work out the area of the triangle?
 The triangle is not drawn to scale because it's not isosceles.
 The statement must be true because the sides are getting longer.
 n can't be 180 because then the angle would not exist.
 You cannot work out the area of the triangle unless n = 90. Then the angle is 90 degrees and the height and base are both 90.
In the first phase of the inquiry, the teacher will introduce the formula and, if appropriate, show how it is derived from the simpler formula (half the product of base and height). As students explore the area, they realise that the contention in the prompt is false. The greatest area is achieved when n = 131.1 (accurate to one decimal place), at which point the area starts to decrease. Students realise that the area must be a minimum at the limits of n. That is because:
 As n tends to 0, the angle gets closer to 180 degrees, and
 As n tends to 180, the angle gets closer to zero degrees.
As you approach both limits, the triangle tends towards a straight line. One way to show this (and the maximum area) is to use a spreadsheet.
In the another phase of the inquiry, students have changed the parameters in the prompt to see how the value of n changes for the maximum area. For example, the side lengths could be (n  1) and the angle (180  10n) degrees and so on. Perimeter inquiry After the initial phase of the inquiry, year 10 students at Haverstock School (London, UK) developed another line of inquiry when they posed questions about the perimeter:
 As n increases, does the perimeter increase?
 What is the perimeter of the triangle when the area is at its maximum?
Michael Joseph, their mathematics teacher, took the opportunity to teach the students about the cosine rule. He created this spreadsheet (using radians) to show that the perimeter is at a maximum when n = 2.61308 (accurate to five decimal places) at which point the angle is (p  2.61308) radians or 30.28 degrees. The prompt could be presented in a simpler form by making the angle smaller, giving students fewer cases to explore. The area increases until n = 61.7, whereupon it starts to decrease. At n = 61.7, the angle is 28.3 degrees and the area equals 902.4 square units (accurate to one decimal place).
 Proof using differentiation and iteration Shawki Dayekh, a teacher of mathematics in north London (UK), proved the maximum area occurs when n = 131.14 (accurate to 2 decimal places). He used differentiation and iteration, expressing the angle in radians. The value of x is 131.14, which means an angle of 48.86 degrees gives the maximum area.
