A vertical pole consists of two portions, the lower being 1/3 of the whole. If the upper
portion subtends an angle Tan' 1/2 at a point in a horizontal plane through the foot of the
plane and distance 40 ft. from it, then the height of the pole is
Answers
Answer:
40 or 120 ft
Step-by-step explanation:
A vertical pole consists of two portions, the lower being 1/3 of the whole. If the upper
portion subtends an angle Tan' 1/2 at a point in a horizontal plane through the foot of the
plane and distance 40 ft. from it, then the height of the pole is
Let say Pole Height = 3P ft
Lower Portion = (1/3)3P = P ft
upper Portion = 3P -P = 2P ft
Horizontal Distance = 40 ft
Angle Subtended by bottom of upper potion = α
Tan α = P/40
Angle Subtended by top of upper potion = β
Tan β = 3P/40
Angle subtended by upper portion = β - α
β - α = Tan⁻¹(1/2)
=> Tan(β - α ) = 1/2
=> (Tanβ - Tanα)/(1 + Tanβ*Tanα) = 1/2
=> (3P/40 - P/40)/( 1 + (3P/40)(P/40) = 1/2
=> 4P/40 = 1 + 3P²/1600
=> 160P = 1600 + 3P²
=> 3P² - 160P + 1600 = 0
=> 3P² - 120P - 40P + 1600 = 0
=> 3P(P - 40) -40(P - 40) = 0
=> (3P - 40)(P - 40) = 0
=> P = 40/3 or 40
Height of pole = 3P = 40 or 120 ft
Answer:
20 and 60m
Step-by-step explanation:
Now tanβ=
60
h
, tan(α+β)=
20
h
tan(α+β)=
1−tanαtanβ
tanα−tanβ
=
20
h
=>
1−
120
h
2
1
+
60
h
=
20
h
120−h
60+2h
=
20
h
h
2
−80h+1200=0
By solving this we get h = 20m or 60m.