The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will
Answers
Answer:
THEN THE ANGLE OF ELEVATION OF ITS TOP WILL BE LESS THAN 60°.
EXPLANATION:-
ACCORDING TO QUESTION.
\tan30 \degree = \frac{h}{x}tan30°=
x
h
= > \frac{1}{ \sqrt{3} } = \frac{h}{x}=>
3
1
=
x
h
= > h = \frac{x}{ \sqrt{3} }=>h=
3
x
now, when height of tower is DOUBLED, we get:
\tan \theta = \frac{2h}{x}tanθ=
x
2h
\tan\theta = \frac{2}{x} \times \frac{x}{ \sqrt{3} }tanθ=
x
2
×
3
x
\tan\theta = \frac{2}{ \sqrt{3} }tanθ=
3
2
\theta < 60 \degreeθ<60°
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The given angle of elevation =30
o
.
Let the height of the tower =h and the viewer be at a distance of x from the foot of the tower.
Then,
x
h
=tan30
o
=
3
1
........(i)
If the height of the tower is doubled then the new height =2h.
Let the angle of elevation of the top be θ.
Then, tanθ=
x
2h
=2×
3
1
=
3
2
......(from i)...........(ii)
But if the angle of elevation doubles then it should be =θ=2×30
o
=60
o
.
Then, tanθ=tan60
o
=
3
........(iii).
Comparing (ii) & (iii), there is a contradiction.
∴ The assertion is incorrect.