if from the top of a tower 50m high,the angles of depression of two objects due north of the lower are respectively 60° and 45° then find the distance between objects 1•50+√2-2) 2•50(√3-3) 3•31 m
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Answer:
Let the two objects be C, D and the height of the tower is 50m
In ΔABC
tan60
0
=
BC
AB
=
x
50
∴x=
3
50
... {i}
Now, In ΔABD
tan45
0
=
BD
AB
⇒1=
x+y
50
⇒x+y=50
⇒y=x−50
⇒y=50−
3
50
...[putting the value from {i}]
⇒y=
3
50
(
3
−1)
∴ Distance between two objects is =
3
50
(
3
−1)
Rationalising the denominator we get the Distance=
3
50
3
(
3
−1)
m
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