Tricky trigonometric word problem

Quote:

Originally Posted by juliajank

I've tried three different approaches to it, and it's taken me well over an hour and I CAN'T figure it out.

The pictures pretty much just for the diagram but in case you can't read the text.

19. The Great Pyramid of Khufu has a square base with a side length of about 230m. The four triangular faces of the pyramid are congruent and isosceles. The altitude of each triangular face makes an angle of 52 degrees. with the base. Find the measure of each base angle of the triangular faces to the nearest degree.

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Those altitudes all meet directly above the center of the square base so dropping a perpendicular from the tip of the pyramid to the base, drawing a perpendicular from the center to the base to the center of one side of the base, and then from that point directly to the tip of the pyramid gives a right triangle in which one leg has length $230/2= 115m$ and angle 52 degrees. The length of the altitude is given by sin(52)= 115/x so x= 115/sin(52). Now look at that triangular face. The altitude, one edge from the base to the tip, and half the base also form a right triangle. If the "base angle of the triangle is $\theta$ , then we have $tan(\theta)= (115/sin(52))/115)= 1/sin(52)$ so $\theta= arctan(1/sin(52))= arctan(csc(52))$ .