# 1. an aeroplane flying horizontally at a height of 1.4km above the ground is observed at an elevation od 60degree. after 15 s , its elevation at the same point is obsreved to be 30degree. calculate the speed of the aeroplane in km/h.2. the angle of elevation of a jet fighter from point A on the ground is 60degree. after a flight of 15 sec, the angle of elevation changes to 30degree. if the jet is flying at a speed of 720km/h, find the constant height at which the jet is flying. (root3=1.732)3. at a point on the ground level, the angle of elevation of the vertical tower is found to be such that its tangent is 5/12. on walking 240m towards the tower, the tangent of the angle of elevation becomes 3/4. find the height of the tower.3.the angles od depressio of the top an dbottom of 8m tall building from the top of a multistorey building are 30 and 45 degrees respectyively. find the height of the multi-storey bldg and the distance b/w the 2 bldgs.4. from the top of a hill. the angles of depression of 2 consecutive km stones due east are found to be 30 and 45 respectively . find the height of the hill.5. a bird sitting on the top of a tree, which is 80m high.the angle of elevation of the bird from a point on the ground is 45. the bird flies away from the point of observation horizontally and remains at a constant height. after 2 sec, the angle of elevation of the bird from the point of observation becomes30. find the speed of the bird.6. the angles of elevation of the top of a tower as seen from 2 points A and B situated in the same line and at distances p and q respectively from the foot of the tower are complementary. prove that the height of the tower is rootpq.7.from a window(60m above the ground) of a house in the stret, the angle of elevation and depression of the foot of the house on the opposite side of the street rae 60 and 45 respectively, show that the height of the opposite house is 60(i = root3).

5)  A bird sitting on the top of a tree, which is 80m high.the angle of elevation of the bird from a point on the ground is 45. the bird flies away from the point of observation horizontally and remains at a constant height. after 2 sec, the angle of elevation of the bird from the point of observation becomes30. find the speed of the bird.

SOLUTION 5)

Let P and Q be the two positions of the bird and let A be the point of observation. Let ABC be the horizontal  line through A.

Given: The angle of elevations of the bird in two positions P and Q from point A are 45° and 30°.

∴ ∠PAB = 45°, ∠QAB = 30°.

Also,

PB = 80 meters In  ΔABP, we have, In  ΔACQ, we have, ∴ PQ = BC = AC – AB = Thus,

The bird touch in 2 seconds.

Hence,

Speed of the bird  6) The angles of elevation of the top of a tower as seen from 2 points A and B situated in the same line and at distances p and q respectively from the foot of the tower are complementary.

prove that the height of the tower is rootpq.

SOLUTION 6) In this figure CD is the tower. Let ∠CAD = θ

7∴ ∠CBD = 90° – θ  (Complementary of θ) 7) From a window(60m above the ground) of a house in the stret, the angle of elevation and depression of the foot of the house on the opposite side of the street rae 60 and 45 respectively, show that the height of the opposite house is 60(i = root3).

SOLUTION 7)  • 1

Let the distance between the two houses is xm.

Let the height of other house if y+60

tan450=60/x

x=60m

Tan600=y/60

y=60root3

height of other house is 60+60root3=60(1+root3) Hence Proved

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