Hello, john doe!
We need more information.
What is that quadrilateral on the hypotenuse . . . a square?
Is that NE-SW slanted line through the center of the square?
. . And is it parallel to the sides of the square?
Assuming all that is true, we have this diagram: Code:
*
* *
A * * *
| * * *
| * * O *
| * * * *
| * /| *
| * /θ| *
| P* | *
| / Q* *
| / | *
| | *
| | *
| | 30° *
* * * * * * * * * * * * *
C R B
We have right triangle $\displaystyle ABC$ with $\displaystyle \angle C = 90^o,\;\angle B = 30^o$
A square sits on hypotenuse $\displaystyle AB$; $\displaystyle O$ is the center of the square.
. . Actually, it could be any point in the square!
$\displaystyle OP \perp AB,\;\;OQR \perp CB$
In right triangle $\displaystyle QRB,\;\angle B = 30^o \quad\Rightarrow\quad \angle BQR = 60^o$
Then $\displaystyle \angle OQP = 60^o$ .(vertical angles)
Therefore, in right triangle $\displaystyle OPQ,\;\theta \:=\:\angle POQ \:=\:30^o$