Find Speed of a Glider

[greatwhiteshark]

From a glider 200 feet above the ground, two sightings of a stationary object directly in front are taken 1 minute apart. The sighting is an angle of elevation of 10 degrees and the second angle of elevation is 40 degrees. What is the speed of the glider?

 

[soroban]

Hello, greatwhiteshark!

As stated, the problem doesn't make sense,
but I think I've figured out what was <u>meant</u>.

From a glider 200 feet above the ground, two sightings of a stationary object

directly in front are taken 1 minute apart.

The sighting is an angle of elevation of 10^o and the second angle of elevation is 40^o.

What is the speed of the glider?

The object cannot be "directly in front".
. . If it was, the angles of elevation would 0^o.

If the pilot had to look up to see the object, the 200-foot altitude is meaningless.
. . [He might as well be taxiing on the ground.]

I believe the object is 'directly in front' but on the ground,
. . and those angles are angles of depression.

        A     x     B     y     D
        * - - - - - * - - - - - +       The object is at C.
        |   * 10d     *40d      :       The glider was at A.
        |       *       *       :
    200 |           *     *     : 200    One minute later,
        |               *   *   :           it is at B.
        |                   * * :
    - - + - - - - - - - - - - - * -     Let AB = x (feet)
                                C       Let BD = y.

. . . . . . . . . . . . . . . . . . . . . . . . . . . .200 . . . . . . . . . . . . 200
In right triangle BDC: . tan 40^o .= .----- . . ---> . . y = ---------- . [1]
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .y . . . . . . . . . . . . tan 40^o

. . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 . . . . . . . . . . . . . .200
In right triangle ADC: . tan 10^o .= .------- . . ---> . . y .= .---------- - x . [2]
. . . . . . . . . . . . . . . . . . . . . . . . . . . .x + y . . . . . . . . . . . . tan 10^o

. . . . . . . . . . . . . . . . . . .200 . . . . . . . .200
Equate [1] and [2]: . ---------- . = . ---------- - x
. . . . . . . . . . . . . . . . . .tan 40^o . . . . tan 10^o

. . . . . . . . . . . . . . . . . 200 . . . . .200
Solve for x: . x . = . ---------- - ---------- . = . 895.905645
. . . . . . . . . . . . . . . .tan 10^o . .tan 40^o


Hence, in one minute, the glider flew 895.9 feet.

In one hour, it would fly: .60 x 895.9 = 53,754 feet.
. . . That equals: .53,754 ÷ 5280 .= .10.18 miles


Therefore, the glider's speed is about 10.18 mph

[As always, someone check my work ... please!]

 

[tkhunny]

I'm a little amazed that it lost no altitude in one minute, but whatever. I guess that's possible.

Definitely unclear description. Rational translation and accurate solution.

 

[greatwhiteshark]

Yes...

Yes, you are right in saying that the angles are depression not elevation. It was a typing error on my part. I like your steps. I will pratice similar questions.