MHB Flagpole Angle: 40 Degrees - What's 3/4 of the Way Up?

  • Thread starter Thread starter oldfart1
  • Start date Start date
  • Tags Tags
    Angle Degrees
AI Thread Summary
The discussion centers on calculating the angle subtended by a flagpole at 3/4 of its height, given that the full height subtends an angle of 40 degrees. The relationship between the height of the flagpole and the distance from the observer is established using the tangent function. By substituting the known angle into the equation, the angle at 3/4 height can be expressed as tan(q) = (3/4) * tan(40 degrees). This allows for the calculation of the angle q using a calculator. The conclusion reached is that the angle at 3/4 of the height is approximately 32 degrees.
oldfart1
Messages
3
Reaction score
0
a flagpole subtends a angle of 40 degrees.
What angle does it subten 3/4 of the way up.
Answer is supposedly 32 degrees ,but how ?

oldfart
 
Mathematics news on Phys.org
Let $$y$$ be the height of the flagpole. Let $$x$$ be the distance from the base of the flagpole to the observer.
Then $$\tan(40^\circ)=\dfrac yx$$. Letting $$q$$ be the angle in question we have $$\tan(q)=\dfrac34\cdot\dfrac yx$$
Can you continue?
 
greg1313 said:
Let $$y$$ be the height of the flagpole. Let $$x$$ be the distance from the base of the flagpole to the observer.
Then $$\tan(40^\circ)=\dfrac yx$$. Letting $$q$$ be the angle in question we have $$\tan(q)=\dfrac34\cdot\dfrac yx$$
Can you continue?

Sorry but no. thanks
 
oldfart said:
Sorry but no. thanks

As Greg pointed out you have $$\tan(q) = \dfrac{3}{4} \cdot \dfrac{y}{x}$$ together with $$\tan(40^{\circ}) = \dfrac{y}{x}$$

That means you can put in $$\tan(40^{\circ})$$ (from the second equation) in place of [math]\dfrac{y}{x}[/math] from the first.

This substitution gives you $$\tan(q) = \dfrac{3}{4} \cdot \tan(40^{\circ})$$

The right hand side of that equation is simply a number your calculator can find for you. Does that make it clearer?
 
Thanks for your help..I get it
Oldfart
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top