Question

1) from the top of a tower 100 m in height a man observes two car on opposite side of the tower with angle of depression 30 degree and 60 degree respectively. find the distance between the car?

2) A jet fighter plane is flying horizontally at speed of 720 km/h. from a point A on the ground, the angle of elevation of the plane is 60 degree. after flight of 15 second the angle of elevation changes to 30 degree. find the height at which the jet is flying

Answer 1

let  the  distance  of  one car  from tower = x,  depression = 30

tan 60  =  x/ 100  or  x = 100 root3 = 173.2 m

let  the  distance of  second  car = y, depression = 60

tan 30 =  y / 100  or  y =  100 / root3  =  173.2 / 3 = 57 .7 m

distance  between cars = 230. 9 m

Answer 2

Solutions 

We all know that from the top of a tower 100 m in height a man observes two car on opposite side of the tower with angle of depression 30 degree and 60 degree respectively. Find the distance between the car? 

Calculations 

[tex]tan 30 = x / 100 tan 60 = y / 100 100 x tan(30) + 100 x tan(60)= (400/3) x sqrt(3)[/tex]  

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A jet fighter plane is flying horizontally at speed of 720 km/h. from a point A on the ground, the angle of elevation of the plane is 60 degree. after flight of 15 second the angle of elevation changes to 30 degree. Find the height at which the jet is flying?

Calculations

[tex]BC = 720 /3600 * 15 = 3 km tan(60) = h/x tan(30) = h / ( x + 3) {h = (3/2)*sqrt(3), x = 3/2} = (3/2)*sqrt(3) = 2.6 km [/tex]