The angle of elevation from a fixed point on the
ground to the top of a building is 30°. If the height
of the building is thriced, then the angle of elevation
of its top from the same point will be
(2) Remains same
(1) Thriced
(3) Doubled
(4) Four times
and how?
Answers
Step-by-step explanation:
Answer:
10 m
Step-by-step explanation:
Refer the attached figure
Height of building = Ab
Height of tower = CD= 50 m
The angle of elevation of the top of a building from the foot of the tower is 30° i.e. ∠ACB =30°
The angle of elevation of the top of the tower from the foot of the building is 60° i.e. ∠DBC =60°
We are required to find the height of building i.e. AB
In ΔDBC
tan\theta = \frac{Perpendicular}{Base}tanθ=
Base
Perpendicular
tan\theta = \frac{DC}{BC}tanθ=
BC
DC
tan60^{\circ}= \frac{30}{BC}tan60
∘
=
BC
30
\sqrt{3}= \frac{30}{BC}
3
=
BC
30
BC= \frac{30}{\sqrt{3}}BC=
3
30
In ΔABC
tan\theta = \frac{Perpendicular}{Base}tanθ=
Base
Perpendicular
tan\theta = \frac{AB}{BC}tanθ=
BC
AB
tan30^{\circ}= \frac{AB}{ \frac{30}{\sqrt{3}}}tan30
∘
=
3
30
AB
\frac{1}{\sqrt{3}}= \frac{AB}{ \frac{30}{\sqrt{3}}}
3
1
=
3
30
AB
\frac{1}{\sqrt{3}}= \frac{\sqrt{3}AB}{30}
3
1
=
30
3
AB
\frac{30}{\sqrt{3}\times\sqrt{3} }= AB
3
×
3
30
=AB
\frac{30}{3}= AB
3
30
=AB
10= AB10=AB
Thus the height of building is 10 m
Answer:
the angle of elevation will be doubled
Step-by-step explanation:
here 'L' is the height of the tower and d is the distance of the fixed point from the foot of the tower
tan 30 =1/√3
therefore,L/d =1/√3
so when we multiply 3 to both sides the
length of the tower becomes 3L(which is requirement
and we get 3L/d as √3
we know that tan 60= √3
therefore our new angle for 3L is 60
