# A tower AB leans towards west making an angle α with the vertical. The angular elevation of B, the topmost point of the tower is β as observed from a point C due west of A at a distance d from A. If the angular elevation of B from a point D due east of C at a distance 2d from C is γ, then prove that         2 tan α = - cot β + cot γ.

Jitender Pal
8 years ago
Hello Student,
Let AB be the tower leaning towards west making an angle α with vertical
At C, ∠of elevation of B is β and at D the
∠ of elevation of B is γ
When in ∆ ABH
⇒ tan α = AH/h ⇒ AH = h tan α . . . . . . . . . . . . . . . . . . . . . . . . . (1)
In ∆ BCH, tan β = h/CH ⇒ CH = h cot β
⇒ d – AH = h cot β
⇒ d = h (tan α + cot β) . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
(Using eqn (1))
In ∆ BDH, tan γ = BH/HD ⇒ h/AH + d
⇒ AH + d = cot γ
⇒ d = h (cot γ – tan α) . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
(Using eqn (1))
Comparing the values of d from (2) and (3), we get
h(tan α + cot β) = h (cot γ – tan α)
⇒ 2 tan α = cot γ – cot β Hence Proved
ALTERNATE SOLUTION :
KEY CONCEPT:
m : n theorem : In ∆ ABC where point D divides BC in the ratio m : n and ∠ ADC = θ
(i) (m + n) cot θ = n cot B - m cot C
(ii) (m + n) cot θ = m cot α – n – cot β
In ∆ BCD, A divides CD in the ratio 1 : 1 where base ∠’s are β and γ and ∠ BAD = 90° + α
∴ By applying m : n theorem we get
(1 + 1) cot (90° + α) = 1. Cot β – 1. cot γ
⇒ -2 tan α = cot β – cot γ
⇒ 2 tan α = cot γ – cot β
Hence Proved.

Thanks
Jitender Pal