Two poles of equal height stand vertically opposite to each other on either side of the road, which is 100 m wide. From a point on the road between the poles, the angle of elevation of the tops of the poles are 30 degree and 60 degree . Find the heights of the poles. Also find the distance of the point from the feets to the poles. Surya don't answer
Answers
Answer:
Given :
Bottom distance = 100 m.
Angle of Elevation \bf{\angle_{1}}∠
1 = 60°
Angle of Elevation \bf{\angle_{2}}∠
2 = 30°.
To find :
Height of the two poles.
Distance of the points from the feet of the poles.
Solution :
Let the height of both the poles will be h m.
Let the distance from point A and B be x m.
Hence according to the Question , the distance from point B will be (100 - x) m.
Height of the tower :
To find the height of pole (in terms of h) with respect to angle 60°.
Using tan θ and substituting the values in it, we get :
\begin{gathered}:\implies \bf{tan\:\theta = \dfrac{P}{B}} \\ \\ \\\end{gathered}
:⟹tanθ=
B
P
\begin{gathered}:\implies \bf{tan\:60^{\circ} = \dfrac{h}{x}} \\ \\ \\\end{gathered}
:⟹tan60∘ = xh
\begin{gathered}:\implies \bf{\sqrt{3} = \dfrac{h}{x}}\quad[\because \bf{tan\:60^{\circ} = \sqrt{3}}] \\ \\ \\\end{gathered}
:⟹ 3 = xh
[∵tan60 ∘ =3 ]
\begin{gathered}:\implies \bf{x = \dfrac{h}{\sqrt{3}}} \\ \\ \\\end{gathered}
:⟹x= 3 h
\begin{gathered}\boxed{\therefore \bf{x = \dfrac{h}{\sqrt{3}}}} \quad \quad Eq..(i) \ \\ \\ \\\end{gathered}
∴x= 3h
Eq..(i)
Hence the distance between base of A and B (in terms of h) is √3/h
Now , by using the tan θ and substituting the values in it, we get :
\begin{gathered}:\implies \bf{tan\:\theta = \dfrac{P}{B}} \\ \\ \\\end{gathered}
:⟹tanθ= BP
\begin{gathered}:\implies \bf{tan\:30^{\circ} = \dfrac{h}{100 - x}} \\ \\ \\\end{gathered}
:⟹tan30 ∘ = 100−xh
\begin{gathered}:\implies \bf{\dfrac{1}{\sqrt{3}} = \dfrac{h}{100 - x}}\quad[\because \bf{tan\:30^{\circ} = \dfrac{1}{\sqrt{3}}}] \\ \\ \\\end{gathered}
:⟹ 31 = 100−xh
[∵tan30 ∘ =31 ]
\begin{gathered}:\implies \bf{\dfrac{100 - x}{\sqrt{3}} = h} \\ \\ \\\end{gathered}
:⟹ 3100−x =h
\begin{gathered}:\implies \bf{\dfrac{100 - x}{\sqrt{3}} = h} \\ \\ \\\end{gathered}
:⟹ 3100−x =h
\begin{gathered}:\implies \bf{100 - x = h\sqrt{3}} \\ \\ \\\end{gathered}
:⟹100−x=3h
Now , by substituting the value of x from equation (i) , we get :
\begin{gathered}:\implies \bf{100 - \dfrac{h}{\sqrt{3}} = h\sqrt{3}} \\ \\ \\\end{gathered}
:⟹100−3h =3h
\begin{gathered}:\implies \bf{\dfrac{100\sqrt{3} - h}{\sqrt{3}} = h\sqrt{3}} \\ \\ \\\end{gathered}
:⟹3100
3−h =h 3
\begin{gathered}:\implies \bf{100\sqrt{3} - h = h\sqrt{3} \times \sqrt{3}} \\ \\ \\\end{gathered}
:⟹100
3
−h=h
3
×
3
\begin{gathered}:\implies \bf{100\sqrt{3} - h = 3h} \\ \\ \\\end{gathered}
:⟹100
3
−h=3h
\begin{gathered}:\implies \bf{100\sqrt{3} = h + 3h} \\ \\ \\\end{gathered}
:⟹100
3
=h+3h
\begin{gathered}:\implies \bf{100\sqrt{3} = 4h} \\ \\ \\\end{gathered}
:⟹100
3
=4h
\begin{gathered}:\implies \bf{\dfrac{100\sqrt{3}}{4} = h} \\ \\ \\\end{gathered}
:⟹
4
100
3
=h
\begin{gathered}:\implies \bf{25\sqrt{3} = h} \\ \\ \\\end{gathered}
:⟹25
3
=h
\begin{gathered}\boxed{\therefore \bf{h = 25\sqrt{3}}} \\ \\ \\\end{gathered}
∴h=25
3
Hence the Height of two towers is 25√3 m.
Distance from the points :
Distance between A and B :
Since, we have taken the base distance as x and we know the value of x in terms of h i.e,
\begin{gathered}:\implies \bf{x = \dfrac{h}{\sqrt{3}}} \\ \\ \\\end{gathered}
:⟹x=
3
h
Now, putting the value of h in the above equation , we get :
\begin{gathered}:\implies \bf{x = \dfrac{25\sqrt{3}}{\sqrt{3}}} \\ \\ \\\end{gathered}
:⟹x=
3
25
3
\begin{gathered}:\implies \bf{x = 25 m} \\ \\ \\\end{gathered}
:⟹x=25m
\begin{gathered}\boxed{\therefore \bf{x = 25\:m}} \\ \\ \\\end{gathered}
∴x=25m
Hence, the base distance from A to B is 25.
Distance between B and C :
We know that the distance between B and C is (100 - x) m.
So by putting the value of x in it , we get :
\begin{gathered}:\implies \bf{100 - x} \\ \\\end{gathered}
:⟹100−x
\begin{gathered}:\implies \bf{100 - 25} \\ \\\end{gathered}
:⟹100−25
\begin{gathered}:\implies \bf{75} \\ \\\end{gathered}
:⟹75
\begin{gathered}\boxed{\therefore \bf{75\:m}} \\ \\ \\\end{gathered}
∴75m
Answer:
QUESTION
Two poles of equal height stand vertically opposite to each other on either side of the road, which is 100 m wide. From a point on the road between the poles, the angle of elevation of the tops of the poles are 30 degree and 60 degree . Find the heights of the poles. Also find the distance of the point from the feets to the poles
ANSWER
In △AOC,
tan60° = h/x , i.e = h / √3
In △BOD,
tan30° = h / 100 -x
1/√3 = h/100-x
100-x = h√3
100- h√3 = x
x = h/√3
therefore ,. 100- h√3 = h/√3
h = 25√3 m
x= h/√3 = 25√3 /√3
x=25 m
∴OC=25 m and OD=75 m.
Step-by-step explanation:
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