A cable car 100m above the ground is seen to have an angle of elevation of 65 degress when it is on a bearing of 345 degress. After a minute, it has an angle of elevation of 69 degress and is on a bearing of 025 degress. Find how far it travels in that minute, and its speed in m.s^-1

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Let A be the position of cable car at 100m above the ground.

Let B the position where observer is standing. and O is point just below the the cable car

Thus AOB is right angle , and angle B is 65 degree.

OA=100 m

Thus

`tan(65^o)=(AO)/(BO)`

`BO=(AO )/tan(65^o)=100/tan(65^o)`

`=46.63 m`

Let A' be after a minute and O' be the point just below the cable car

Thus

`tan(69^o)=(A'O')/(BO')`

`BO'=(A'O')/(tan(69^o))=100/tan(69^o)`

`=38.39m`

distance travelled by cable car in 1 minute=BO-B'O'

=46.63-38.39

=8.24 m

Thus speed of the cable car=`8.24/60=0.14 ms^(-1)`

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