TRIGONOMETry Problems (Patulong)

[THF]

Q:
At a certain point, the angle of elevation of the top of a flagpole which stands on level ground is 40 degrees. 80 feet nearer the pole, the angle of elevation is 65 degrees. How high is the pole?

[dummy_c]

Pwede kaya 80tan65?? Pero parang hindi kasi parang easy lang. I don't understand the first statement.

[???]

Ratio and proportion lang yan. Two right triangle.

1. Where at 40 degree distance from base of the pole (assuming horizontal plane at x axis) lets say that point from pole base is mark as point "r". And pole base is 0.

2. And 80 ft nearer the pole seeing the highest point of the pole its angular degree is 65 degree. Mark as point " x".
                                    p
                                    I
                                    I
                                    I
                                    I
        / 40 d       /65d       I

       r               x           0( base)

3. where distance from r to x = 80 ft.  Point r = x + 80ft = adjacent side
    ratio and proportion on right triangle    point x, 0, p then x. And right triangle point r,0,p then r.

Just solve the other unknown. Goodluck.
         

[kinyo]

h = (80*sin(40)*sin(65))/sin(25)

law of sines

 

[???]

In simple term you must first solve the unknown "x" to get the true height of the pole. Height commonly denote as "y" is subtituted by the angular displacement. Using your fav SOH CAH TOA method. Since your talking trigonometry.

[motion55]

tan 65° = y/x
tan 40° = y/(x+80)

You now have 2 linear equations with 2 unknowns. It is straight algebra to solve this.

 

[THF]

Tama kaya tong ginawa ko?

let: y1 be the height in 40 degrees
      y2 be the height in 65 degrees

Solution:

y1=(x+80)tan40
y2=xtan65

y1 must equal y2 (y1=y2)

xtan65  =  (x+80)tan40

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[Johannah Aleksandria]

Very simple problem actually, ayoko lang mag-isip . basta ts pag-aralan mo sign law ska cosine law.

[melecee]

may isang ganyang-ganyan problem na lumabas sa nakalipas na board ng ee april 17-18 sa math.

[iyo karpo]

hcot65 + 80 = hcot40

h = 58.056

[/]

nice post..

[Nokia]

first equation: tan 40 = [h / (80+x)]
second equation: tan 65 = (h/x)
xtan65  =  (x+80)tan40
using shift solve,
x = 51.423 deg
substituting the value of x in the first or second equation gives
h= 110.27 deg

[NORTH LUZON]

112.797 ft