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Hello,

My question concerns the task 14 from the 2015 high school final exams. Task content:
The base of the \(ABCDS\) pyramid is the square \(ABCD\). The side edge \(SD\) is the height of the
pyramid, and its length is twice the length of the base edge. Calculate the sine of the angle between the \(ABS\) and \(CBS\) sidewalls of this pyramid.

I solved the problem correctly, but in the end after the trigonometric one, when two sines come out, the negative one is rejected. The question is why the angle alpha is in the range \((0^\circ; 90^\circ)\) and the sine cannot be negative?

In the pyramids, the side walls are isosceles acute-angled triangles. This is basically enough to explain why sin is only positive

The angle you are looking for is obtuse, but his sines is still positive!

I greet You

[edited]

my draw