Suppose, there are two windows in a house. A window of the house is at a height of 1.5 m
above the ground and the other window is 3 m vertically above the lower window. Anil and
Sanjeev are sitting inside the two windows. At an instant, the angles of elevation of a
balloon from these windows are observed as 45° and 30°, respectively.
(i) Find the height of the balloon from the ground.
(ii) Among Anil and Sanjeev, who is more closer to the balloon?
(iii) If the balloon is moving towards the building, then will both the angles of elevation
remain same?
Answers
Answer:
1) 6.23 m
2) Sanjeev (upper window)
3) No, it will change constantly
Step-by-step explanation:
Let the height of ballon over the upper window be x and distance between the building and balloon be B
In 1st scenario the distance between the building and baloon is constant.
As, angle of elevation from lower window is 45.
Taking tan45 = x+3/B
B = x + 3 (tan 45= 1)
As, angle of elevation from upper window is 30.
Taking tan30 = x/B
B = (root3)x, (tan 30 = 1/(root3))
So, (root3)x = x + 3
Solving this, we get x = (root3)
So Total height of building is 1.73 + 3 + 1.5
= 6.23m
2) The point which makes a smaller angle for another point is always closer than the point which makes higher angle.
3) As the object moves closer, the angle of elevation will always keep changing.
Answer:
Given, two windows are at 1.5 m & 4.5 m from ground respectively.
AG=1.5 m,AB=3 m
⇒GB=4.5 m
Let OX=h(height of balloon from ground)
∴QX=(h−4.5)
Two perpendicular are dropped from A to B respectively to P & Q.
In right triangle APX
tan45=
AP
PX
⇒
AP
(h−4.5+3)
=1
⇒AP=(h−1.5)
∴BQ=(h−1.5)
So, in right triangle, BQX
tan30=
BQ
QX
=
3
1
⇒
h−1.5)
(h−4.5)
=
3
1
⇒
3
h−
3
(4.5)=(h−1.5)
⇒(
3
−1)h=
3
(4.5)−1.5
⇒h= 3
−1
3
(15)(3)−1.5
=8.6 m.