Two sides of a triangle are of lengths 2a and 2b
and contain an angle of 120°. If the angle opposite
the side 2a is θ, tan θ is equal to
(a) a√2 /a + 2b
(b) a √3 /2a + b
(c b√ 3 /a + 2b
(d) b √3 /2a + b
62first find the oppsoite side (3rd unknown side)=x
Then if the angle is theta...
then sinθ/2a=sin120/x
also cos θ = x2+(2b)2-(2a)2
2xb
now I think you can get tan theta
106let the opposite side be x...
so sinθ/2a=sin120/x ==> x2 = 3a2/sin2θ ... (i)
now cos120 = [(2a)2 + (2b)2 - x2]/(2*2a*2b)
==> -1/2 = (4a2+4b2-x2)/8ab
==> -4ab = 4a2+4b2-x2
==> x2 = 4a2+4b2 + 4ab ... (ii)
equating (i) and (ii),
sin2θ = 3a2/(4a2+4b2 + 4ab)
==> tan2θ = 3a2/(a2+4b2+4ab) = 3a2/(a+2b)2
==> tanθ = a√3/(a+2b)
