Question

13. The SMHS Surveying Team decided to go on a vacation to a small island in the South Pacific.They chartered plane 2135 through the "Blue Ribbon Rent-A-Good Plane" Company. As the planeapproached the island, which was 1600 feet wide, the pilot, realizing that if you were on the SMHS surveying team that you were an excellent math student, announced that the angle of depression to the near side of the island was 25° and the angle of depression to the far side of the island was 22°. AT THAT VERY MOMENT the pilot realized that the plane was in DEEP TROUBLE and he must at that very instant know the altitude of the plane!!! The team, having studied under an exceptional math teacher, found the altitude of the plane and saved the day!!What was the plane's altitude?

Answer 1

First, Lte's draw a picture of our problem:

which is equivalent to the following picture:

where h denotes the altitute and x is the horizontal distance from the near side of the island to the airplane.

Now, from triangle ACD, we have that

`tan68=(1600+x)/(h)`

By distributing h, this last equation is equivalent to

`tan68=(1600)/(h)+(x)/(h)\text{ ...\lparen A\rparen}`

Now, from triangle BCD,we have

`tan65=(x)/(h)`

By substituting this result into equation (A), we have

`tan68=(1600)/(h)+tan65`

Then, by moving tan65 to the left hand side, we get

`tan68-tan65=(1600)/(h)`

or equivalently,

`(1600)/(h)=tan68-tan65`

By taking reciprocals, we get

`(h)/(1600)=(1)/(tan68-tan65)`

so, by moving 1600 to the right hand side, we get

`h=(1600)/(tan68-tan65)`

which gives

`h=(1600)/(2.47508-2.14450)`

Then, the altitude is given by

`h=4839.9792\text{ ft}`

Therefore, by rounding to the nearest whole number, the answer is 4840 feet.