$ABC$
with $\angle A=\frac{360}{k}$
degrees, where $k\ge5$
is an integer and $\angle B=90$
degrees, as follows: the first triangle with $A$
at $(0,0)$
and $B$
at $(0,1)$
, then repeatedly place triangles with $A$
at $(0,0)$
such that $AB$
of the new triangle coincides with $AC$
of the previous triangle. What is the size of the area enclosed by the first $k$
triangles?

 Donald starts off with an empty mailbox. If Donald receives
$2$
emails on the first day, and the amount of emails he receives per day triples compared to the day before, how many days would it take for him to have over$10000$
emails?  Alice wants to know the height of a lamp post. As the lamp post is very tall, she cannot measure this directly. Instead, she measures the length of the shadow, which is
$10$
metres long. Knowing that the$6$
foot tall bus stop sign nearby casts a$4$
foot shadow, how tall is the lamp post? What is the angle between the tip of the shadow and the top of the lamp post? How does this angle compare with the angle of the bus stop?
 Donald starts off with an empty mailbox. If Donald receives

What relationship do you notice between the 2nd and 1st triangle?

More specifically, between the sides but more importantly between the areas of these triangles?

What about any two successive triangles?

What is the area of the first triangle?

The total area is the sum of the areas of all triangles. Notice anything familiar about that sum given the above relationship?

More specifically, given that the ratio between the areas of any two successive triangles is constant, what kind of sum is the sum of all their areas?

Let
$\theta=\frac{360}{k}.$
The first triangle has area$\frac{\tan\theta}{2}.$
After placing the next triangle we notice that it is similar to the previous triangle: the ratio between sides is$\frac{1}{\cos \theta}$
and the ratio between areas is$1/\cos^2 \theta.$
This applies to all pairs of successive triangles, i.e. a geometric progression, hence the total area is the geometric series:$$ \begin{aligned} A&=\frac{\tan \theta}{2}\sum_{i=0}^{k1} \frac{1}{\cos^{2i} \theta} \\&=\frac{\tan \theta}{2}\cdot\frac{1/\cos^{2k} \theta 1}{1/\cos^2 \theta 1}, \end{aligned} $$
which can be simplified to$$ \begin{aligned} A&=\frac{\tan \theta \cos^2 \theta}{2}\cdot\frac{1/\cos^{2k} \theta 1}{\sin^2 \theta} \\&=\frac{1}{2 \tan \theta} \cdot \left( \frac{1}{\cos^{2k} \theta} 1 \right). \end{aligned} $$