Question 64783
<pre>
In a triangle we have <font face = "symbol">Ð</font>A=60°, <font face = "symbol">Ð</font>B=80° and <font face = "symbol">Ð</font>C=40°
we draw median, angle bisector and altitude from vertex A to 
opposite side(BC).  what are the value of all angles that 
are constructed on vertex A.
<font size = 5><b>
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Let D be the midpoint of BC. Then AD is the median. 

AB/sin40° = BC/sin60° = AC/sin80°

To make things easier, let
p = sin40° = .6427876097
q = sin60° = .8660254038 
r = sin80° = .984897753

AB/p = BC/q = AC/r

AB = BC·p/q

AC = BC·r/q

We calculate median AD using the law of cosines on <font face = "symbol">D</font>ACD

AD² = AC² + CD² - 2·AC·CD·cos40°

substitute CD = BC/2 and AC = BC·r/q

AD² = BC²r²/q² + BC²/4 - 2·BC·r/q·BC/2·cos40°

AD² = BC²r²/q² + BC²/4 - BC²r·cos40°/q

Factor out BC² on the right:

AD² = BC²(r²/q² + 1/4 - r·cos40°/q)
         ________________________  
AD = BC·<font face = "symbol">Ö</font>r²/q² + 1/4 - r·cos40°/q
         ________________________ 
Let k = <font face = "symbol">Ö</font>r²/q² + 1/4 - r·cos40°/q = .8197650971

So

AD = BC·k

By the law of sines in triangle ACD

CD/sin(<font face = "symbol">Ð</font>CAD) = AD/sin40²

CD/sin(<font face = "symbol">Ð</font>CAD) = AD/p

sin(<font face = "symbol">Ð</font>CAD) = CD·p/AD
                                  
Substitute CD = BC/2 and AD = BC·k
                           
sin(<font face = "symbol">Ð</font>CAD) = (BC/2)·p/(BC·k)

the BC's cancel and we have
                      
sin(<font face = "symbol">Ð</font>CAD) = p/(2k) = .6427876097/(2·.8187650971)
 
     = .3920559755

<font face = "symbol">Ð</font>CAD = 23.08248883°

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Let E be the point where the bisector of <font face = "symbol">Ð</font>A
intersepts BC.

Then <font face = "symbol">Ð</font>CAE = 30° because that's half of 60°

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Draw altitude AF perpendicular to BC. <font face = "symbol">D</font>CAF is
a right triangle and so <font face = "symbol">Ð</font>CAF is complementary
to <font face = "symbol">Ð</font>C, which is 40° and so

<font face = "symbol">Ð</font>CAF = 50° 

Now you have three of the angles at A.
You can easily find any of the other angles
at A by simple addition and subtraction.

Edwin</pre>