Question

the angle of elevation of top of tower from to point at distance of 4 metre and 9 metre from base of tower and in the same straight line with it are complementary prove that height of tower is 6 metre

Answer 1

Given AB is the tower.
P and Q are the points at distance of 4m and 9m respectively.
From fig, PB = 4m, QB = 9m.
Let angle of elevation from P be α and angle of elevation from Q be β.
Given that α and β are supplementary. Thus, α + β = 90
In triangle ABP,
tan α = AB/BP – (i)
In triangle ABQ,
tan β = AB/BQ
tan (90 – α) = AB/BQ (Since, α + β = 90)
cot α = AB/BQ
1/tan α = AB/BQ
So, tan α = BQ/AB – (ii)
From (i) and (ii)
AB/BP = BQ/AB
AB^2 = BQ x BP
AB^2 = 4 x 9
AB^2 = 36
Therefore, AB = 6.
Hence, height of tower is 6m.

Answer 2

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$\huge{ \mathfrak{here \: is \: answer}}$


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$\huge \boxed{hope \: it \: helps}$



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