# SOLUTION: <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 al

Algebra ->  Algebra  -> Triangles -> SOLUTION: <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 al      Log On

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 Geometry: Triangles Solvers Lessons Answers archive Quiz In Depth

 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(8882)   (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. ---------------------------------------------- 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° --------------------- Let E be the point where the bisector of ÐA intersepts BC. Then ÐCAE = 30° because that's half of 60° ---------------------- 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```