SOLUTION: In triangle ABC, c = 10 and angle A = angle B = 40o. Find the length of the median to line segment AC

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Question 43248: In triangle ABC, c = 10 and angle A = angle B = 40o. Find the length of the median to line segment AC
Found 2 solutions by psbhowmick, venugopalramana:
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
In triangle ABC, BD is the median which bisects the side AC.
Drop a perpendicular from C on AB intersecting AB at E.



In triangle ABC, < CAB = < CBA = 40%5Eo and AB = 10 units.
As triangle ABC is isoscles, so the perpendicular dropped from the vertex on the opposite bisects the opposite side.
Hence, AE = BE = AB%2F2 = 10%2F2 = 5 units.

Now, in triangle AEC, < AEC is right angle.
So, AE%2FAC = cos(< CAE)
or 5%2FAC+=+cos%2840%5Eo%29
or AC+=+5%2Fcos%2840%5Eo%29
or AC+=+5%2F0.839
or AC+=+6.527

So, AD = 1%2F2AC = 6.527%2F2 = 3.263 units.

Now, consider triangle ADB.
Apply cosine formula
BD%5E2+=+AD%5E2+%2B+AB%5E2+-+2%2AAD%2AABcos(< BAD)
or BD%5E2+=+3.263%5E2+%2B+10%5E2+-+2%2A3.263%2A10%2Acos%2840%5Eo%29
or BD%5E2+=+60.655
or BD+=+sqrt%2860.655%29
or BD = 7.788 units

Thus, the length of the median to the line segment AC is 7.788 units.

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
In triangle ABC, c = 10 and angle A = angle B = 40o. Find the length of the median to line segment AC
let the median from B to AC be BE=M
AB=10...ANGLE A = ANGLE B =40...HENCE...AC=BC=X SAY...AND ANGLE C=180-40-40=100
10/SIN(100)=X/SIN(40)
X=10*SIN(40)/SIN(100)=6.523=AC=BC
IN TRANGLE BEC
EC=AC/2=3.262
BC=6.523
ANGLE BCE =100
BE^2=BC^2+CE^2-2*BC*CE*COS(BCE)=6.523^2+3.262^2-2*6.523*3.262*COS(100)
BE^2=60.538
BE=7.78