The angles of elevation of a flying kite at three points A,B,C on a horizontal line (lying in the same vertical plane with the kite) are in the ratio 1 : 2 : 3. If AB=a, BC=b, then the height of the kite is

detailed solution

Correct option is A

Let Q=kite, A=first point observation, B=second point observation, C= third point of observation.  Given AB =a, BC=b.  Let PQ = h. Since exterior angle = sum of the opposite interior angles,∠PBQ=∠BQA+∠BAQ    and ∠PCQ=∠CBQ+∠CQB∴AB=a=QBApplying sine rulein ΔBQC, we get  BQsin∠QCB=BCsin∠CQB⇒asin1800−3θ=bsinθ⇒4sin2θ=3−ab=3b−ab ⇒sin2θ=3b−a4b∴cos2θ=1−sin2θ=1−3b−a4b=4b−3b+a4b=b+a4bFrom PBQ, sin2θ=ha⇒2sinθcosθ=ha ⇒4sin2θcos2θ=h2a2⇒4.3b−a4b.a+b4b=h2a2 ⇒h2=a24b23b−aa+b∴h=a2b3b−aa+b