In equilateral triangle ABC let points D and E trisect BC. Find sin DAE.
It would be much easier if I knew how to draw on here, but I don't. So here goes. You should take a sheet of paper and draw as we go.
Draw the equilateral triangle. Label the vertexes. Mark your points D and E to trisect BC. Draw AD and AE.
Draw a perpendicular from angle A to the center of BC. Label the point on BC as P.
On my drawing, it looks like if you could get the sine of angle DAP and double it, you'd have the sine of DAE. I'm not certain of my footing about this assumption.
Assign a numerical value to each side AB, AC, and BC. It doesn't matter what the value is since we're going to be working with proportions, so why not call it 6 so at least some of the numbers will come out even.
You can figure the length of AP because it's the height of the original triangle.
You can figure the length of DP because it's half of DE which was a third of BC.
Use Pythagorean Theorem to figure the length of AD.
The sine of DAP is DP divided by AD.
Double the sine of DAP and you've got the sine of DAE. I think.
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