in tri.abc,angle a = 60.prove that

Dear Student,

 

Consider the figure below

 

 

In triangle ACD,

 

AD/AC = cos60

 

=> AD/AC = 1/2

 

=> AD = AC/2

 

Also,

 

CD/AC = sin60

 

=> CD/AC = sqrt(3)/2

 

=> CD = AC x sqrt(3)/2

 

 

Now, in triangle CDB,

 

BC² = CD² + BD² = CD²  + (AB - AD)²

 

=> BC² = AC² x 3/4 + (AB² + AD² - 2xABxAD)

 

=> BC² = 3/4 x AC² + AB² + 1/4 x AC² - ABxAC

 

=> BC² = AC² + AB² - ABxAC

 

Hence Proved.

 

Regards Topperlearning.