Two poles of equal heights are standing opposite each other on either side of the roads, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.
Answers
Step-by-step explanation:
The values we have,
AB = 80 cm
AC = BD = h
AP = x and BP = 80 - x
→ Perpendicular/Base = tan∅
→ AC/AP = tan 30°
→ AC/x = 1/√3
→ AC = x/√3
→ BD/BP = tan 60°
→ BD/(80 - x) = √3
→ BD = √3(80 - x)
Since, AC = BD
→ x/√3 = √3(80 - x)
→ x = 240 - 3x
→ 4x = 240
→AP = x = 60
So, BP = (80 - 60) = 20
→ AC = 60/√3
→ AC = 60√3/3
→ AC = BD = 20√3
Answer
Height of both poles is 20√3 m and point is 60 m away from left pole and 20 m away from right pole.
Now .....the figure is included with the attachment.....
Let ,the distance of the point p from the pole AB is =X metres
And AB=CD=h (let)
And from the pole CD is =(80-x) metres
Now ....tan60°=AB/x=h/x
=>>x=AB/tan60°
=>>x=h/tan60°..........(i)
Now ....tan30°=CD/(80-x)
=>>x=80-[CD/tan30°]
=>>x=80-[h/tan30°]........(ii)
From .... equations..(i) and (ii)....
80-[h/tan30°]=h/tan60°
=>>h[{1/tan60°}+{1/tan30°}]=80
=>h[√3+(1/√3)]=80
=>>h[(3+1)/√3]=80
=>>h=(80√3)/4
=>>h=20√3. Metres
Therefore the height of the poles is 20√3 metres....
Now ..,the distance of the point from the pole AB is =20√3/√3=20metres
And from pole CD is
=(80-20)=60metres
