14. An aeroplane at a height of 600 m passes vertically above another aeroplane at
an instant when their angles of elevation at the same observing point are 60° and 45°
respectively. How many meters higher is the one from the other
Answers
ANSWER
Let the aeroplanes are at point AA and DD respectively. Aeroplane AA is flying 600600 m above the ground.
So, AB=600AB=600
\angle ACB=60^o,\angle DCB=45^o∠ACB=60
o
,∠DCB=45
o
From \triangle ABC△ABC, we have
\cfrac{AB}{BC}=\tan 60^o
BC
AB
=tan60
o
\Rightarrow BC=\cfrac{600}{\sqrt{3}}=200\sqrt{3}⇒BC=
3
600
=200
3
From \triangle DCB,△DCB,
\cfrac{DB}{BC}=\tan 45^o\Rightarrow DB=200\sqrt{3}
BC
DB
=tan45
o
⇒DB=200
3
Since, the distance AD = AB -BD = 600-200\sqrt{3}AD=AB−BD=600−200
3
AD =200(3-\sqrt{3})=200(3-1.7321)=253.58AD =200(3−
3
)=200(3−1.7321)=253.58 m
Hence, the distance between the two aeroplanes is 253.58253.58 m.
Answer:
The distance between the two aeroplanes is 253.58 m.
Explanation:
Let's consider that point A and point D are where the aeroplanes are placed.
AB=600
∠ACB=60°
∠DCB=45°
From △ABC, we have
AB/BC= tan 60°
BC=600/√3=200√3
From ,△DCB,
DB=200√3
AD = AB -BD = 600-200√3
=200(3-1.7321)=253.5 m
Hence, the distance between the two aeroplanes is 253.58 m.
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