The angles of elevation of the top of a tower from two points on the ground at distances 9 m and 4 m from the base of the tower are in the same straight line with it are complementary. Find the height of the tower.
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SOLUTION:
Let AB = h m be the height of the Tower.
Let ∠ACB = θ
∠ADB = (90°- θ) [complementary means the sum of two angles is 90°]
In ΔABC
tan θ = AB/ BC= P/B
tan θ = h/4..,..….........(1)
In ΔABD
tan (90°- θ) = AB/BD
cot θ = h/9 ... .…......(2) [tan(90°- θ)=cot θ]
On multiplying eq (1) and (2)
h/9 × h/4 = tan θ ×cot θ = 1
[tan θ .cot θ =1]
h²/36 = 1
h² = 36
h =√36
h = ± 6 m
h = 6 m [height is always positive]
Hence, the height of the tower is 6 m.
HOPE THIS WILL HELP YOU...
Let AB = h m be the height of the Tower.
Let ∠ACB = θ
∠ADB = (90°- θ) [complementary means the sum of two angles is 90°]
In ΔABC
tan θ = AB/ BC= P/B
tan θ = h/4..,..….........(1)
In ΔABD
tan (90°- θ) = AB/BD
cot θ = h/9 ... .…......(2) [tan(90°- θ)=cot θ]
On multiplying eq (1) and (2)
h/9 × h/4 = tan θ ×cot θ = 1
[tan θ .cot θ =1]
h²/36 = 1
h² = 36
h =√36
h = ± 6 m
h = 6 m [height is always positive]
Hence, the height of the tower is 6 m.
HOPE THIS WILL HELP YOU...
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