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In triangle ABC, angle BAC is 120 degrees.  D is on BC so that AD bisects angle BAC.  If AB = 2*AC and AD = 100, then find BC.

 Jan 10, 2020
 #1
avatar+109739 
0

Since   BAC  is a bisected apex angle....then

 

AB / BD  =  AC / CD

2AC / BD  = AC / CD

2AC /AC = BD/ CD

2 = BD / CD

2CD = BD

 

And by the Law of Cosines  we have that

 

(BD)^2  =  AD^2  + (AB)^2  - 2 (AD*AB)cos (60)

(2CD)^2 = AD^2  +  (2AC)^2- 2 (AD * 2AC) (1/2)

4CD^2  = AD^2  + 4AC^2 - 2(AD*AC)      (!1)

 

And

(CD)^2 =  AD^2  + AC^2  - 2(AD*AC)cos (60)

CD^2  =  AD^2 + AC^2 - (AD*AC)     multiply through by 4

4CD^2 = 4AD^2 + 4AC^2 - 4(AD*DC)    (2)

 

Set  (1)   = (2)

 

AD^2 + 4AC^2 - 2(AD*DC)  = 4AD^2 + 4AC^2 - 4(AD*DC)

3AD^2  - 2(AD*DC)    = 0

AD ( 3AD - 2DC)  = 0

3AD - 2DC   =0

3 (100)  = 2DC

DC  = 150

 

And   2CD  = BD  =  300

 

So

 

BC  =  BD + DC  = 300 + 150    = 450

 

 

cool cool cool

 Jan 11, 2020
 #2
avatar+531 
+3

∠ BAC = 120°             The law of cosines:      a² = b² + c² - 2ab * cosA  

AC = 150                         (BC)² = (AB)² + (AC)² - 2(AB)(AC) * cos(BAC) 

AB = 300                                                        BC = 396.86      indecision

Dragan  Jan 11, 2020
edited by Dragan  Jan 11, 2020
edited by Dragan  Jan 11, 2020
edited by Dragan  Jan 11, 2020

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