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

Asked by suryap2 | 25th Sep, 2010, 10:38: AM

Expert Answer:

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.

Answered by  | 25th Sep, 2010, 11:57: AM

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