the angle of elevation of the top of a tower from two points at a distance of 4m and 9m from the base of the tower and in the same straight line with it are 60°30° respectively.find the height of the tower.
Answers
Answered by
11
HERE IS YOUR ANSWER :-
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.
MARK AS BRAINLIEST
THANX
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.
MARK AS BRAINLIEST
THANX
Anonymous:
Jab angle de rkhe h to alpha beta kyo laga rahe ho????
ok
Answered by
2
Answer:9 root 3
Hope it will help you
Attachments:
Similar questions