Asked by rahul kumar
If the sides of ∆ABC are 4,5 6 then prove that largest angle will be twice of smallest angle
Answers
Answered by
Reiny
6^2 = 4^2 + 5^2 - 2(4)(5)cosØ
cosØ = (16+25-36)/(40) = 1/8
Ø = 82.819..
4^2 = 5^2 + 6^2 - 2(5)(6)cos A
cosA = (25+36-16)/60 = 3/4
A = 41.4096..
and 2(41.4096..) = 82.81924... same as Ø
or ... an actual "proof"
from above:
cos(smallest angle) = cosA = 3/4
cos(largest angle) = 1/8
we have to show that cos 2A = 1/8
cos 2A = 2cos^2 A - 1 , one of our main identities
= 2(9/16) - 1
= 9/8 - 1
= 1/8
cosØ = (16+25-36)/(40) = 1/8
Ø = 82.819..
4^2 = 5^2 + 6^2 - 2(5)(6)cos A
cosA = (25+36-16)/60 = 3/4
A = 41.4096..
and 2(41.4096..) = 82.81924... same as Ø
or ... an actual "proof"
from above:
cos(smallest angle) = cosA = 3/4
cos(largest angle) = 1/8
we have to show that cos 2A = 1/8
cos 2A = 2cos^2 A - 1 , one of our main identities
= 2(9/16) - 1
= 9/8 - 1
= 1/8
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