Question

the triangle ABC is given. angle bisector of the angle CAB intrsect side BC in point D. if angle ADB 45 and CD.BD =AD^2 find angle BAC . solve the problem by using law of sine

Answer 1

Given : triangle ABC is given. angle bisector of the angle CAB intrsect side BC in point D

angle ADB 45 and CD.BD =AD²

To Find  angle BAC

Solution:

Let say ∠BAC = 2a

=> ∠CAD = ∠BAD = a

∠ADB = 45°

∠ADB =  ∠ACD + ∠CAD

=> 45° = ∠ACD + a

=> ∠ACD  = 45° - a

∠ADB = 45°=>  ∠ADC = 180° - 45° = 135°

∠ADC =   ∠ABD + ∠BAD

=> ∠ABD =  135° - a

in ΔADB

AD/Sin(135 - a)   = BD/Sina

in ΔADC

AD/Sin(45 - a)   = CD/Sina

AD²/Sin(135 - a)Sin(45 - a)  = BD.CD/sina.sina

=> AD²sin²a = BD.CDSin(135 - a)Sin(45 - a)

∵ AD² = BD.CD

=> sin²a  = Sin(135 - a)Sin(45 - a)

=> 2sin²a  = 2Sin(135 - a)Sin(45 - a)

2 sinAsinB = cos(A − B) − cos(A + B)

=> 2sin²a  = Cos(90) - Cos(180 - 2a)

=> 2sin²a  = 0  +  Cos2a

=> 2sin²a = 1 - 2sin²a

=> 4sin²a = 1

=> sin²a = 1/4

=> sina = 1/2

=> a = 30°

=> 2a = 60°

∠BAC =  60°

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