Derive the value of sin60 geometrically
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the ans is in the picture.
:-)hope it helps u.
:-)hope it helps u.
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Answered by
195
hello...
firstly u draw an equilateral triangle
ABC in which Draw a perpendicular AD .
BD = a and DC = a and AB = 2a.
Now in ∆ABD
AB^2 = Ad^2 + BD ^ 2
( 2a )^ 2 = AD ^ 2+ a ^2
4a^2 - a ^2 = AD ^2
3a ^2 = AD^2
√3a = AD
Then ,
sin 60= p/h
= √3a/ 2a
= √3/ 2
since , Sin60 = √3/2
hence proved ......
please mark it as a brainieist answer......
firstly u draw an equilateral triangle
ABC in which Draw a perpendicular AD .
BD = a and DC = a and AB = 2a.
Now in ∆ABD
AB^2 = Ad^2 + BD ^ 2
( 2a )^ 2 = AD ^ 2+ a ^2
4a^2 - a ^2 = AD ^2
3a ^2 = AD^2
√3a = AD
Then ,
sin 60= p/h
= √3a/ 2a
= √3/ 2
since , Sin60 = √3/2
hence proved ......
please mark it as a brainieist answer......
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