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Two stations due south of a leaning tower which leans towards the north are at distances a and b from its foot. If α , β be the elevations of the top of the tower from these stations, prove that its inclination ? is given by cot ? = b cot α - a cot β/b-a .
Ans: Let AE = x, BE = h
Tan φ =BE/AE =h/x
x= h x (1/tan φ)
x = h cot φ ---------------1
tan α =BE/CE = h /a + x
a + x = h cot α
x= h cot α - a --------------2
tan β = BE/DE = h/b + x
b+x = h cot β
x = h cot β - b ---------------3
from 1 and 2
h cot φ = h cot α - a
h ( cot φ + cot α ) = a
h = a /- cot φ + cot α -------------4
from 1 and 3
h cot φ = h cot β - b
h ( cot φ - cot β) = b
h= b/- cot φ + cot β
from 4 and 5
a/- cot φ + cot α = b/- cot φ + cot β
a (cot β - cot φ ) = b ( cot α - cot φ )
- a cot φ + b cot φ = b cot α - a cot β
(b - a) cot φ = b cot α - a cot β
cot φ = b cot α - a cot β/b - a
Prove the subsequent Boolean expression: (x∨y) ∧ (x∨~y) ∧ (~x∨z) = x∧z Ans: In the following expression, LHS is equal to: (x∨y)∧(x∨ ~y)∧(~x ∨ z) = [x∧(x∨ ~y)] ∨ [y∧(x∨
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