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Consider triangle ABC with AB=5, BC=6. Let D be a point on AC so that CD=2(AD) and angle CBD equals two times angle ABD. Find the length of side AC.

Thank you Apr 30, 2021

#1
+3

ABD = x

CBD = 2x

DC = 2y

BD = h

Note that BAD * 2 = BCD.

2*5*sine(x)*h = 6*h*sine(2x)

10*sine(x) = 6*sine(2x)

5/3sine(x) = sine(2x)

5/3sine(x) = 2*cos(x)*sine(x)

5/6 = cos(x)

x = 33.5573098

3x = 100.6719294

AC^2 = 5^2 + 6^2 - 5*6*cos(100.6719294 degrees)

AC = 8.15815883728

That problem took way too much effort, please tell me if it's correct and if there's a better solution. :))

=^._.^=

Apr 30, 2021
#3
+2

Hello, catmg!

I like your method; good job!!! There're only a couple of errors:

Note that BAD * 2  BCD.                  ( ∠ABD * 2 = ∠CBD )

AC^2 = 5^2 + 6^2 - 2*5*6*cos(100.6719294 degrees)

AC = 8.491826135

civonamzuk  May 1, 2021
#4
+1

Ohh, thank you.

I accidentally forgot the 2. :))

=^._.^=

catmg  May 1, 2021
#2
+2

This one is a little sticky  !!!

Using the Law of Sines  twice we have  :

sin ( CBD) / CD =  sin CDB  /   BC               ( sin ADB  = sin CDB)   (  CBD  = 2 ABD )  ( CD = 2AD)

Equating (1)  and (2)   we  have

3sin *( 2 sin ABD cos ABD)  =  5 sin ABD

6 cos ABD  =  5

cos ABD =  5/6

sin ABD  =  sqrt  ( 1 - 25/36)  = sqrt (11)  / 6

cos  CBD =  cos ( 2 ABD)  =  1 - 2sin^2 ( ABD)  = 1  -  2 * 11 / 36   =  7/18

sin CBD =  sqrt ( 1 - (7/18)^2 )   = 5sqrt (11)  / 18

cos  ( ABC)  = cos ( ABD + CBD)  =

cosABDcosCBD  - sin ABDsinCBD =

(5/6)(7/18)  - sqrt (11) / 6  *  5sqrt (11)  / 18   =  -5/27

Using the Law of Cosines

AC^2  =  AB^2  +   BC^2 -  2 ( AB * BC)  cos ( ABC)

AC^2  =   5^2  +  6^2  - 2 ( 5 * 6) ( -5/27)

AC^2   =  649 / 9

AC =  sqrt (649) / 3    ≈  8.49

I think catmg   just made a slight error in using the Law of Cosines......otherwise....our answers  would  agree

One thing for sure.....catmg's is  WAY  less complicated   !!!!   May 1, 2021
edited by CPhill  May 1, 2021
edited by CPhill  May 1, 2021