Question 64783:
In a triangle we have ÐA=60°, ÐB=80° and ÐC=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.
Answer by Edwin McCravy(20059)   (Show Source):
You can put this solution on YOUR website!
In a triangle we have ÐA=60°, ÐB=80° and ÐC=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.
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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 DACD
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·Ör²/q² + 1/4 - r·cos40°/q
________________________
Let k = Ör²/q² + 1/4 - r·cos40°/q = .8197650971
So
AD = BC·k
By the law of sines in triangle ACD
CD/sin(ÐCAD) = AD/sin40²
CD/sin(ÐCAD) = AD/p
sin(ÐCAD) = CD·p/AD
Substitute CD = BC/2 and AD = BC·k
sin(ÐCAD) = (BC/2)·p/(BC·k)
the BC's cancel and we have
sin(ÐCAD) = p/(2k) = .6427876097/(2·.8187650971)
= .3920559755
ÐCAD = 23.08248883°
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Let E be the point where the bisector of ÐA
intersepts BC.
Then ÐCAE = 30° because that's half of 60°
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Draw altitude AF perpendicular to BC. DCAF is
a right triangle and so ÐCAF is complementary
to ÐC, which is 40° and so
ÐCAF = 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
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