Pre-Calc Help?

Let the equal sides of the triangle be of length s

Let the vertex angle between those sides be θ

Drop a perpendicular from the vertex to the base resulting in two right triangles. The vertex angle is halved resulting in two angles that are each θ/2.

From that we can find the length of the perpendicular (aka the height of the triangle)

h = s cos(θ/2)

We can also find the length of the base which is:

b = 2 s sin(θ/2)

The area of the triangle is:

A = ½bh

A = ½(2 s sin(θ/2) * s cos (θ/2))

A = ½ s² (2 * sin(θ/2) * cos(θ/2))

Using the double angle formula

2 sin(θ/2) * cos(θ/2) = sin(2(θ/2)) = sin θ

So the formula for the area reduces down to:

A = ½ s² sin θ

I'll leave the other steps (drawing a labeled sketch and creating two examples of using the formula) for you to finish.