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
+2401 ABD = x
CBD = 2x
AD = y
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. :))
=^._.^=
+1694 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
+128406 This one is a little sticky !!!
Using the Law of Sines twice we have :
sin ( ABD) / AD = sin ADB / AB
sin (ABD) /AD = sin ADB / 5
sin (ABD) = AD sin ADB / 5
5 sin (ABD) / AD = sin ADB (1)
sin ( CBD) / CD = sin CDB / BC ( sin ADB = sin CDB) ( CBD = 2 ABD ) ( CD = 2AD)
sin (2ABD) / (2AD) = sn ADB / 6
sin (2ABD) = (2AD) sin ADB / 6
sin (2ABD) = AD sin ADB / 3
3* sin (2ABD)) / AD = sin ADB (2)
Equating (1) and (2) we have
3sin ( 2ABD) / AD = 5sin (ABD) / AD
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 !!!!
