Two poles of equal heights are standing opposite each other on either side of the road, 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.
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Answers
Question:
Two poles of equal heights are standing opposite each other on either side of the road, 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.
Answer:
Let AB and CD be the poles of equal
height
O is the point between them from where
the height of elevation taken.
BD is the distance between the poles.
Alq,
AB = CD,
OB + OD = 80 m
Now,
In right triangle CDO,
tan 30º = CD/OD
=> 1/43 = CD/OD
CD = OD/V3... (1)
also,
In right triangle ABO,
tan 60° = AB/OB =
= AB/(80-OD) » V3
AB = v3(80-OD)
AB = CD (Given)
v3(80-OD) = OD/v3
3(80-OD) = OD
→ 240 - 3 OD OD
- 4 OD = 240
- OD = 60
Putting the value of OD in equation (1) CD = OD/v3 = CD = 60/v3 = CD = 20v3 m =
also,
OB + OD = 80 m → OB = (80-60) m = 20 m Thus, the height of the poles are 2013 m and distance from the point of elevation are 20 m and 60 m respectively.
Answer:
20√3
Step-by-step explanation:
Given AB=ED=h , BD=80
Let the given point be C such that BC=x and CD=80−x
In ΔABC
tan30°= AB/BC
= 1/√3 = h/x
⇒x= √3h
In ΔECD
tan60°= ED/CD
=√3= h/80-x
⇒h=(80-x) √3
Put the value x=√3h, we get
h=(80−√3h) √3
⇒h=80√3-3h
⇒4h=80√3
⇒h=20√3
Then, height of poles h=20√3
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