36none??????
ur approach...................??
q) In a triangle ABC, AC = BC then sin [3(A + B)4] equals...
i) Sin {B + 2C2}
ii) Sin {A + 2C2}
iii) Sin {B + 4C2}
iv) Sin (B - 3C)
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6 Answers
1if AC = BC the angle B= angle A
=> first 2 options are same. are you sure you have typed the question correctly?
36options are perfectly right...!!
a bit tricky bt nt much...!!
actually i remembered a question of the same sort...
so designed it myself... nd quite sure for the options now as well...
i'll post the solution..
actually just wanted to see what approach do people have towards this question..!!
a bit tricky nd not a quality question at all
71It could be easily verified also. Set any angle θ and check the options.
36yea.... ofcourse.!!
actually i used the hit and trial method in my phase test in a q from S.O.T......
ii assumed a right-triangle with sides 3,4,5...!!
using it i got R = 1
but the options were
R , 1/R , ..... , .......
i ticked R but unfortunately it was 1/R .... :P
36well, here's the soln.
simply draw the angle bisector of A metting BC at D
LADC = B + B/2 = 3B/2
and, sin LADC = sin (A/2 + C) = sin (B/2 + C) since A = B
=> sin (3B)2 = sin (A + 2C)2 = sin (B + 2C)2
or, sin {3(2B)/2} = sin (A + 2C)2 = sin (B + 2C)2
or, sin 3(A + B)2 = sin (A + 2C)2 = sin (B + 2C)2
so, options (A) and (B).....!!