Question

A triangle has two corners with angles of `pi / 12` and `pi / 12`. If one side of the triangle has a length of `9`, what is the largest possible area of the triangle?

Answer 1

`frac(81)(4)`

Explanation:

To create a triangle with maximum area, we want the side of length `9` to be opposite of the smallest angle. Let us denote a triangle `DeltaABC` with `angleA` having the smallest angle, `frac(pi)(12)`, or `15` degrees. Thus, `BC=9`, and since the triangle is isoceles, `AC=BC=9`. Angle `C` measures `150` degrees (`180-15-15`).
Now, we may find the area of the triangle through the formula `Area=frac(1)(2)ab sin(C)`, or `frac(1)(2)AC*BC sin(C)=frac(81)(4)`.

Answer 2

`81/4=20.25`

Explanation:

Given that two of the angles are `pi/12`, the triangle is isosceles, and the third angle is `pi-pi/12-pi/12=(5pi)/6`

If one side of the triangle has a length of `9` unit, then its maximum area will occur when this length is opposite to one of its smaller angles, as shown in the diagram.

Area of a triangle `A=1/2ab*sinx`,
where `a and b` are two sides of the triangle and `x` is the included angle between the two sides `a and b`.

Here, given the triangle is isosceles, `=> a=b=9`.
`=> A=1/2*a^2*sinx`

`=>` Area of the triangle `A= 1/2*9^2*sin((5pi)/6)=1/2*81*1/2=81/4=20.25`