Two roads intersect at 34 degrees. There is one car on each road, and they travel at different speeds of 80km/h and 100km/h. How far apart are the cars after 2 hours? After 2 hours there is a helicopter flying above the 2 cars, the angle of depression is 20 degrees, and the straight line distance is 100km. Determine the altitude of the helicopter.

After 2 hours the fast car A is `200 \ km` away, and the slow car B is `160\ km` away from the intersection C. The altitude of the helicopter is `34.2\ km`.

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Two roads intersect at `34^{o}`. There are 2 cars traveling on each road at two different speeds of `80` km/hr and `100` km/hr. We have to find out how far apart the 2 cars are after 2 hours. Also given that a helicopter flies above the cars with an angle...

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Two roads intersect at `34^{o}`. There are 2 cars traveling on each road at two different speeds of `80` km/hr and `100` km/hr. We have to find out how far apart the 2 cars are after 2 hours. Also given that a helicopter flies above the cars with an angle of depression `20^o` and a straight line distance of `100` km, we have to find the altitude of the helicopter.

See the attachment for the figure.

Figure A represents the location of the fast car, and B represents the slow car. C is the intersection of the two roads. H represents the helicopter, assuming that it is flying in the middle of the two cars. h is the altitude of the helicopter.

After 2 hours, the fast car A is `100 ` km/hr `\times` `2` hr = `200\ km. `

The slow car B is `80` km\hr `\times ` `2` hr = `160\ km` , from the intersection C.

This is assuming that here the angle of depression `20^o` and the straight line distance`100\ km` is given to the slow car B. We can find the altitude of the helicopter by considering the right angle triangle ` Delta HDB` .

`sin20^o = \frac{h}{100}`

implies `h=100 sin 20^o``= 34.2\ km` .

Therefore the altitude of the helicopter is `34.2\ km` .

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