SOLUTION: Provide a sketch for this information. From a horizontal distance of 10.5km, a pilot observes that that the angles of depression of the top and base of a control tower are 36degre

Algebra ->  Trigonometry-basics -> SOLUTION: Provide a sketch for this information. From a horizontal distance of 10.5km, a pilot observes that that the angles of depression of the top and base of a control tower are 36degre      Log On


   



Question 1142030: Provide a sketch for this information.
From a horizontal distance of 10.5km, a pilot observes that that the angles of depression of the top and base of a control tower are 36degrees and 41degrees respectively. Calculate height of the control tower

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
here's your sketch.

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point A is the airplane.
point F is the top of the tower.
point C is the bottom of the tower.

AB is the horizontal distance from the plane to the tower at the level of the plane.

DC is the horizontal distance from the plane to the tower at the level of the bottom of the tower.

EF is the horizontal distance from the plane to the tower at the level of the top of the tower.

ABCD is the rectangle formed.

angle BAF is the angle of depression from the plane to the top of the tower.

angle BAC is the angle of depression from the plane to the bottom of the tower.

AF forms the diagonal of rectangle ABFE.

AC forms the diagonal of rectangle ABCD.

angle AFE is equal to angle BAF because they form alternate interior angles of parallel lines AB and EF.

angle ACD is equal to angle BAC because they form alternate interior angles of parallel lines AB and DC.

two triangles formed are triangles AFE and ACD.

the base of both of these triangles (EF and DC) are each 10.5 kilometers in length.

the length of BF is equal to y.

the length of FC is equal to x.

the length of BC is equal to x + y.

in triangle AFE, tangent of angle AFE is equal to y / 10.5.

in triangle ACD, tangent of angle ACD is equal to (x + y) / 10.5

since angle AFE is equal to 36 degrees, you get tan(36) = y / 10.5

since angle ACED is equal to 41 degrees, you get tan(41) = (x + y) / 10.5

solve for y in the first equation of tan(36) = y / 10.5 to get:

y = 10.5 * tan(36).

multiply both sides of the second equation of tan(41) = (x + y) / 10.5 to get:

10.5 * tan(41) = x + y

replace y with 10.5 * tan(36) to make that equation become:

10.5 * tan(41) = x + 10.5 * tan(36)

subtract 10.5 * tan(36) from both sides of that equation to get:

10.5 * tan(41) - 10.5 * tan(36) = x

you now have:

y = 10.5 * tan(36).

x = 10.5 * tan(41) - 10.5 * tan(36)..

solve for each to get:

y = 7.628696544.

x = 1.498814203.

that makes x + y = 9.127510747.

the height of the tower is equal to x.

therefore, your solution is that the height of the tower is equal to 1.498814203 kilometers.

sounds awfully high for a tower, but that's what i get from my understanding of the information given.