If the angles of elevations of the top of a tower from three collinear points A,B and C on a  line leading to the foot of the tower are 30∘,45∘ , and 60∘ respectively, then the ratio AB:BC is

Let ' h ' be the height of the tower MN From ΔMCN,tan⁡60=hx⇒x=h cot60°,and from ∆MBN,tan 45=hBN⇒h=BN⇒ BC=h−x⇒BC=h-hcot60° From ΔMAN,tan⁡30=hAB+h⇒AB+h=h cot30°⇒ AB=hcot⁡30∘−h∴ABBC=hcot⁡30∘−1h1−cot⁡60∘=3−11−13=3∴AB:BC=3:1