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In equilateral triangle ABC let points D and E trisect BC. Find sin DAE.

 Aug 18, 2019
 #1
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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.

 

.

 Aug 20, 2019
 #2
avatar+103122 
+1

Let the sides of the equilateral triangle be some convenient integer, say, 6

 

Draw altitude  AF ......this makes DF  = 1

 

Using the Law of Sines we have that

 

AF/ sin 60  =  BA/ sin 90

 

AF  / sin 60  = 6

 

AF   =  6√3 / 2   =   3√3  =  √27  =  

 

And using the Pythagorean Theorem

 

DA  =  √[ AF^2  + DF^2]    =  √[ 27 + 1]    = √28  =  EA

 

And using the Law of Cosines

 

DE^2  =  DA^2  + EA^2   -  2(DA)(EA) cos DAE

 

2^2 = 28 + 28  -  (2*28) cos DAE       simplify

 

4  - 56 =  -56 cosDAE

 

-52  = -56 cosDAE

 

-52/-56  = cos DAE

 

13/14  = cosDAE

 

So....sin DAE  =  +√[ 1 - cos^DAE ]  = + √[ 1 -  (13/14)^2 ]   =  +√ [196 - 169] / 14  =  √27 / 14 =  (3/14)√3 

 

 

cool cool cool

 Aug 20, 2019

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