Question Number 104735 by bemath last updated on 23/Jul/20

Commented by1549442205PVT last updated on 23/Jul/20

what is your question?

Commented bybemath last updated on 23/Jul/20

find the value of sin 18° in  geometrically

Answered by 1549442205PVT last updated on 24/Jul/20

Commented by1549442205PVT last updated on 24/Jul/20

Drawing an isosceles triamgle ABC,CA=CB=a  with ACB^(�) =36°.Then CAB^(�) =CBA^(�) =72°.  A point D taken on the side BC such that  DA=DC=m.Denote the midpoont of   side AB by E.Then we get DCA^(�) =DAC^(�) =36°  ⇒DAB^(�) =72°−36°=36°⇒ADB^(�) =72°,so  ΔABD is isosceles at A⇒AB=AD=m  and hence ΔABD⋍ΔCAB(a.a)  ⇒((AD)/(BC))=((BD)/(AB))⇔(m/a)=((a−m)/m)⇔a^2 −am−m^2 =0  This is a quadratic equation for a with  (√Δ)=(√(m^2 +4m^2 ))=m(√5).Hence  a=((m+m(√5))/2)=((m((√5)+1))/2).  Since CE is the bisector line of ΔABC  ,BCE^(�) =(1/2)ACB^(�_ ) =18° and CE⊥AB,so  sin18°=sinBCE^(�) =((BE)/(BC))=(m/2):a=(m/2):((m((√5)+1))/2)  =(1/((√5)+1))=((1×((√5)−1))/(((√(5+1)))((√5)−1)))=(((√5)−1)/4)  Thus,sin18°=(((√5)−1)/4)  which  we need to find