A man observes a tower AB of height h from a point P on the ground. He moves a distance d towards the foot of the tower and finds that the angle of elevation is doubled. He further moves a distance 3d/4 in the same direction and finds that the angle of elevation is three times that of P.
Prove that $36 h^{2} = 35 d^{2}$ .

Question

A man observes a tower AB of height h from a point P on the ground. He moves a distance d towards the foot of the tower and finds that the angle of elevation is doubled. He further moves a distance 3d/4 in the same direction and finds that the angle of elevation is three times that of P.
Prove that

36 h^{2} = 35 d^{2}

.

Filo · Accepted Answer

Let RB=x∠BQR is ext ∠ of △PBQ∴∠PBQ=2α−α=αNow in △PBQ.∠PBQ=∠QPB⇒PQ=QB=dAlso ∠BRA is ext ∠ of △BQR∴∠BQR=3α−2α=αAnd ∠QBR=π−3α (linear pair)Now in △BQR by applying Sine law, we getd/sin(π−3α)=3d/4sinα=x/sin2α⇒d/sin3α=3d/4sinα=x/sin2α⇒d/3sinα−4sin3α=3d/4sinα=x/2sinαcosα⇒d/3−4sin2α=3d/4=x/2cosα............(I)(II)(III)From eq (I),I=II⇒d/3−4sin2α=3d/4⇒4=9−12sin2α⇒sin2α=5/12⇒cos2α=7/12Also from eq (I) using (II) and (III) we have3d/4=x/2cosα⇒4x2=9d2cos2αx2=9d2/4=7/12=21/16d2.................(3)Again from △ABR we have sin3α=h/x⇒3sinα−4sin3α=h/x⇒sinα(3−4sin2α)=h/x⇒sinα[3−4×5/12]=h/x (using sin2α=5/12)⇒4/3sinα=h/xSquaring both sides we get16/9sin2α=h2/x216/9×2/12=h2/x2(again using sin2α=5/12)⇒h2=4×5/9× 3x2⇒h2=20/27×21/16 d2[using value of x2 from eq (3)]⇒h2=35/36 d2⇒36h2=35 d2Which was to be proved.

Question Text	A man observes a tower AB of height h from a point P on the ground. He moves a distance d towards the foot of the tower and finds that the angle of elevation is doubled. He further moves a distance 3d/4 in the same direction and finds that the angle of elevation is three times that of P. Prove that $36 h^{2} = 35 d^{2}$ .
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