In A ABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn
from point C to side AB
Answers
Answer:
Step-by-step explanation:
use analytic geometry
B(0,0)
A(8,0)
C(x,y)
such that
x^2 + y^2 = 144
(x - 8)^2 + y^2 = 100
C(27/4, 15√7 / 4)
now,
cosB = (12^2 - 8^2 - 10^2)/(2*8*10) = -20/160 = -1/8
cosB/2 = √((1 - 1/8)/2) = √7 / 4
thus
tanB/2 = 3/√7
thus the angle bisector is:
y = 3x/√7
we then get the median:
mdpt of BC (27/8, 15√7 / 8)
y = (15√7/8)/(27/8 - 8) (x - 8)
y = (-15√7)(x - 8)/37
intersecting the lines:
x = 35/9
y = 5√7 / 3 << answer
...Show more
00
Sa T's avatar
Sa T
Lv 5
1 decade ago
OK. There may be properties tying together bisectors and medians but I do not remember them. We shall solve it just by using triangle relationships.
Draw your triangle ABC, AB=8, AC=10 and BC=12.
Draw your median from A down to side BC opposite to cut BC at point D so that BD=DC=6.
Draw your angle ABC bisector to meet line AD at point E.
Drop the normal from E to side AB to cut it at point F.
The distance you are asked to find is FE.
Use Cosine Rule to find angle ABC.
cosABC= (AB^2+BC^2-AC^2) /(2*AB*BC).
Drop the height from A normal on to BC to cut BC at G
Height AG= AB sinABC
CosBAG= AG/AB find and note angle BAG
BG= AB cosABC find and note distance BG
Distance GD=BD-BG=6-BG
tan GAD=GD/AG find and note angle GAD.
Angle BAD= angle BAG+ angle GAD
Using Sine Rule on triangle ABD,
AD/sinABD=BD/sinBAD=AB/sinBDA
The middle term is known so you can find AD and BDA
Angle BED=180-(ABD/2 + BDA)
Sine Rule on triangle ABE,
BE/sinBDA=BD/sinBED from which you find BE
FE= BE sin(ABD/2) = your answer! now you put the numbers in.
...Show more
00