An aeroplane at a altitude of 200 metre above a river observes the angle of depression of opposite points on the two banks or 45 degree and 60 degree find the width of the river in metres
Answers
Hi ,
Draw a rough diagram of given data
|A
|
|
| 60
|__________|___________45_
B. C. D
Join A to C and D
Let
Height of the aeroplane from the
ground = AB = 200m
Width of the river = CD
i) In right triangle ABD,
tan 45° = AB / BD
1 = 200 / BD
BD = 200m ---( 1 )
ii ) In triangle ABC ,
Tan 60° = AB / BC
√3 = 200 / BC
BC = 200 /√3
BC =( 200 √3)/( √3 × √3 )
BC = 200√3 / 3----( 2 )
iii ) width of the river =
CD = BD - BC
= 200 - 200√3/3
= 200( 1 - √3 /3)
= 200( 3 - √3 )/3
=( 200/3) ( 3 - √3 ) m
I hope this helps you.
:)
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Answer:
✯ Given :-
An aeroplane at an altitude of 200 m observes the angles of depression of opposite points on the two banks of a river to be 40° and 60° .
✯ To Find :-
What is the width of the river.
✯ Solution :-
» Let, AD be the height of the aeroplane
» And, BC = x be the width of the aeroplane.
⋆ Given that, AD = 200 m
➟ In ∆ABD,
tan45°=BD/AD
⟹1= BD/AD
⇒ AD = BD
⇒ BD = 200 m
Again,
➟ In ∆ACD
✯ Given :-
An aeroplane at an altitude of 200 m observes the angles of depression of opposite points on the two banks of a river to be 40° and 60° .
✯ To Find :-
What is the width of the river.
✯ Solution :-
» Let, AD be the height of the aeroplane
» And, BC = x be the width of the aeroplane.
⋆ Given that, AD = 200 m
➟ In ∆ABD,
tan45°= BD/AD
⟹1= BD/AD
⇒ AD = BD
⇒ BD = 200 m
Again,
➟ In ∆ACD,
tan60°= CD/AC
⟹ 3 =CD/AC
⇒ BC = BD + CD
⇒ BC = 200 + 115.4
➥ BC = 315.4 m
therefore the width of the river is 315.4m