A flagstaff is placed on top of a building. The flagstaff and building subtend equal angles at a point on level ground which is 200 m away from the foot of the building. If the height of the flagstaff is 50 m and the height of the building is h, which of the following is true?
h3 - 50h2 + (200)2h + (200)250 = 0
None of these
h3 + 50h2 + (200)2h - (200)250 = 0
h3 - 50h2 - (200)2h + (200)250 = 0
Explanation:
Let AD be the flagstaff and CD be the building.
Assume that the flagstaff and building subtend equal angles at point B.
Given that AD = 50 m, CD = h and BC = 200 m
Let $\angle$ABD = θ, $\angle$DBC = θ (∵ flagstaff and building subtend equal angles at a point on level ground).
Then, $\angle$ABC = 2θ
From the right $\triangle$ BCD,
tanθ=$\dfrac{DC}{BC}=\dfrac{h}{200}$ ⋯(eq:1)
From the right $\triangle$ BCA,
tan2θ=$\dfrac{AC}{BC}=\dfrac{AD + DC}{200}=\dfrac{50 + h}{200}$
⇒$\dfrac{2tanθ}{1−tan2θ}=\dfrac{50 + h}{200}$ (∵tan(2θ)=$\dfrac{2tanθ}{1−tan2θ}$)
⇒$\dfrac{2\left( \dfrac{h}{200}\right)}{1−\dfrac{h^{2}}{200^{2}}}=\dfrac{50 + h}{200}$(∵ substituted value of tan θ from eq:1)
⇒2h=$\left(1−\dfrac{h^{2}}{200^{2}}\right) $ (50 + h)
⇒2h=50+h−$\dfrac{50h^{2}}{200^{2}}−\dfrac{h^{3}}{200^{2}}$
⇒2(2002)h =50(200)2+h(200)2−50h2−h3
(∵ multiplied LHS and RHS by 2002)
⇒h3+50h2+(200)2h−50(200)2=0
Let AD be the flagstaff and CD be the building.
Assume that the flagstaff and building subtend equal angles at point B.
Given that AD = 50 m, CD = h and BC = 200 m
Let $\angle$ABD = θ, $\angle$DBC = θ (∵ flagstaff and building subtend equal angles at a point on level ground).
Then, $\angle$ABC = 2θ
From the right $\triangle$ BCD,
tanθ=$\dfrac{DC}{BC}=\dfrac{h}{200}$ ⋯(eq:1)
From the right $\triangle$ BCA,
tan2θ=$\dfrac{AC}{BC}=\dfrac{AD + DC}{200}=\dfrac{50 + h}{200}$
⇒$\dfrac{2tanθ}{1−tan2θ}=\dfrac{50 + h}{200}$ (∵tan(2θ)=$\dfrac{2tanθ}{1−tan2θ}$)
⇒$\dfrac{2\left( \dfrac{h}{200}\right)}{1−\dfrac{h^{2}}{200^{2}}}=\dfrac{50 + h}{200}$(∵ substituted value of tan θ from eq:1)
⇒2h=$\left(1−\dfrac{h^{2}}{200^{2}}\right) $ (50 + h)
⇒2h=50+h−$\dfrac{50h^{2}}{200^{2}}−\dfrac{h^{3}}{200^{2}}$
⇒2(2002)h =50(200)2+h(200)2−50h2−h3
(∵ multiplied LHS and RHS by 2002)
⇒h3+50h2+(200)2h−50(200)2=0