Question

the angle of elevation of the top of tower from two points at a distance of 4m and 9m from the base of tower and in a Same straight line with it are complementary. Prove that the height of the tower is 6m
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Answer 1

Answer:

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. Therefore, AB = 6

Answer 2

Answer:

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.