Two roads intersect at 34 degrees.  Two cars leave the intersection on different roads at speeds of 80 km/h and 100 km/h.  After 2 h, a traffic helicopter which is above and between the two cars takes readings on them.  The angle of depression to the slower car is 20 degrees and the distance to it is 100 km.  How far is the helicopter from the faster car?  The answer is supposed to be (38.7 km) but how do I get that?

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We assume that the helicopter is directly above the line drawn between the cars.

(1) We can find the land distance between the cars: We have a triangle with sides 160,200 and an included angle of `34^@` . (The angle is between the roads -- the lengths come from the...

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We assume that the helicopter is directly above the line drawn between the cars.

(1) We can find the land distance between the cars: We have a triangle with sides 160,200 and an included angle of `34^@` . (The angle is between the roads -- the lengths come from the cars travelling at a constant speed for 2 hours.)

Using the Law of Cosines we can find the distance between the cars:

`d^2=160^2+200^2-2(160)(200)cos34^@`

Thus `d~~112"km"`

Now consider the triangle formed by the two cars and the helicopter. We know two sides and the included angle: 100,112,and `20^@` .

The sides are the distance between the cars that we computed and the distance from the slow car to the helicopter which is given. Since the angle of depression from the helicopter to the car is 20 degrees, the angle of elevation from the car to the helicopter is 20 degrees.

Now use the Law of Cosines again to find the distance to the heliciopter:

`d^2=100^2+112^2-2(100)(112)cos20^@)`

So `d~~38.66"km"`

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