needed very urgentlyyyyyyy
@msingh as always in triangles, the side is opposite the angle - side a is opposite angle A, etc.
k
@msingh still need help with this ?
is the answer sqrt 3 ? this can be done using sine rule. and ofcourse, knowing the fact that A,B,C are in AP
oh, and this too. sin 2x = 2sin x cos x
First of all for solving this question do you know the conception of : Arithmetic Progression of trigonometric equations. And after doing that, I personally guarantee that you can on your own solve this particular question. I personally believe, the happiness of solving Mathematics is much better.
dude, you tried ?
@goformit100 why are you posting this on every question?
can any one solve it plse just want to clear my doubt plseee
i will give you a start, you can try further. A,B,C in A.P ----> A+C = 2B in a triangle, A+B+C = pi find the value of angle B from here.
yes , i will 2B+B= 180 B = 60
yes. so, A+C =.... ?
also, you know sine rule ? what is it ?
A+C= 2*60=120
first find a/c from sine rule. (and c/a) which are ... ?
you know sine rule, right ? or should i write it ?
a/sinA=b/sinB=c/sinC
yes, so, a/c =... ?
srry , i don't know
a/sinA=c/sinC a/c = sin A/ sin C
okay
in your question, we need a/c sin 2C , right ? with sin 2C = 2 sin C cos C what will a/c sin 2C simplify to ??
a/c (2sinCcosC)
we had a/c = sin A/ sin C also. why didn't you put it ?
@agent0smith Sir I am just inspiring the users to solve questions on there own, simply enlightening their inner power of solving mathematics.
(sinA/ sinC )*sin2C =2sin^2A
:O sin A / sin C * 2sin C cos C sin C gets cancelled 2 sin A cos C wake up ?
what about c/a sin 2A ?
oops i m srry for that (sinC/ sinA )*sin2A 2sin C cos A
yes, so you have 2 sin A cos C + 2sin C cos A = 2 (sin A cos C + sin C cos A ) =.... ? can you think of a standard formula ?
just last 2 steps remaining...
2 sin(A + C)
yes,! and you know the value of A+C so, 2 sin (A+C) = 2 sin 120 =.. ?
underroot 3
thats what i also got :)
@hartnn thank u so much bro. really thanks
@msingh welcome so much bro. really welcome ^_^
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