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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