The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds the angle of elevation changes to 30°. If the jet plane is flying at a constant height of
Let A be the point of observation, C and E be the two points of the plane. It is given that after 15 seconds angle of elevation changes from 60° to 30°.
i.e., ∠BAC = 60° and ∠DAE = 30°. It is also given that height of the jet plane is 1500
[Since jet plane is flying at constant height, therefore, CB = ED =
In right triangle ABC, we have
In right triangle ADE, we have
Putting the value of (i) in (ii), we get1500 + BD = 4500⇒ BD = 3000∵ Distance travelled in 15 sec= CE = BD = 3000 metres,
Now, speed of plane (m/s) =
and speed of plane (km/h) =
= 720 km/hr
The height of a tower is 10 m. What is the length of its shadow when Sun's altitude is 45°?
Let BC be the length of shadow is x m
Given that: Height of tower is 10 meters and altitude of sun is 45°
Here we have to find length of shadow.
So we use trigonometric ratios.
In a triangle ABC,
`⇒ tan = (AB)/(BC)`
`⇒ tan 45°=(AB)/(AC)`
`⇒1=10/x`
`⇒x=10`
Hence the length of shadow is 10 m.
Concept: Heights and Distances
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If the ratio of the height of a tower and the length of its shadow is `sqrt3:1`, what is the angle of elevation of the Sun?
Let C be the angle of elevation of sun is θ.
Given that: Height of tower is `sqrt3` meters and length of shadow is 1.
Here we have to find angle of elevation of sun.
In a triangle ABC,
`⇒ tanθ =(AB)/(BC)`
`⇒ tan θ=sqrt3/1` ` [∵ tan 60°=sqrt3]`
`⇒ tan θ=sqrt3`
`⇒ θ=60 °`
Hence the angle of elevation of sun is 60°.
Concept: Heights and Distances
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