At a point D, the angle of elevation of a tower of a lower is such that its tangent is 512; on walking 120 meters nearer the tower the tangent of the angle of elevation is 34. The height of the tower is

It is given that at a point D, the angle of elevation of a tower of a tower is such that its tangent is 512. Now walking 120 meters nearer the tower the tangent of the angle of elevation is 34.Let ∠ADB = ϕ and ∠ACB = θ and DC = 120 m.According to the given data the picture has been drawn below.Here AB is the tower, In ΔABD, tanθ= Perpendicular base tanϕ=ABDB⇒512=ABDB……….(i) Given tanϕ=512On walling 240m near to tower, it reaches point C, then in ΔABC, tanθ= Perpendicular base tanθ=ABBC⇒34=ABBC………(ii) Given tanθ=34Let BC=x m, AB=y m. Now in ΔABC, tanθ=ABBC⇒34=yx (from equation (ii)) ⇒x=43y Now ,tanϕ=ABDB ⇒512=y120+x (from equation (i))⇒y120+43y=512 ⇒12y=600+203y ⇒36y-20y=1800 ⇒16y=1800 ⇒y=180016 ⇒y= 112.5 mSo the height of the tower is 112.5 m.So, the correct option is 1.

At a point D, the angle of elevation of a tower of a lower is such that its tangent is 512; on walking 120 meters nearer the tower the tangent of the angle of elevation is 34. The height of the tower is

It is given that at a point D, the angle of elevation of a tower of a tower is such that its tangent is 512. Now walking 120 meters nearer the tower the tangent of the angle of elevation is 34.Let ∠ADB = ϕ and ∠ACB = θ and DC = 120 m.According to the given data the picture has been drawn below.Here AB is the tower, In ΔABD, tanθ= Perpendicular base tanϕ=ABDB⇒512=ABDB……….(i) Given tanϕ=512On walling 240m near to tower, it reaches point C, then in ΔABC, tanθ= Perpendicular base tanθ=ABBC⇒34=ABBC………(ii) Given tanθ=34Let BC=x m, AB=y m. Now in ΔABC, tanθ=ABBC⇒34=yx (from equation (ii)) ⇒x=43y Now ,tanϕ=ABDB ⇒512=y120+x (from equation (i))⇒y120+43y=512 ⇒12y=600+203y ⇒36y-20y=1800 ⇒16y=1800 ⇒y=180016 ⇒y= 112.5 mSo the height of the tower is 112.5 m.So, the correct option is 1.