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

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Discussion Overview

The discussion revolves around determining the angle subtended by a flagpole at a height of 3/4 of its total height, given that the flagpole subtends an angle of 40 degrees from a certain distance. The scope includes mathematical reasoning and trigonometric relationships.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant states that the angle subtended by the flagpole is 40 degrees and questions what the angle would be at 3/4 of the way up.
  • Another participant introduces variables for the height of the flagpole and the distance from the observer, establishing the relationship using the tangent function.
  • A repeated post reiterates the same mathematical setup, indicating a request for further assistance.
  • A later reply clarifies the substitution of the tangent of 40 degrees into the equation to find the angle at 3/4 height, suggesting that this leads to a calculable value.
  • One participant expresses gratitude for the clarification and indicates understanding of the explanation provided.

Areas of Agreement / Disagreement

The discussion shows some agreement on the mathematical approach but does not reach a consensus on the final angle value or the method of calculation, as some participants seek clarification while others provide assistance.

Contextual Notes

The discussion does not resolve the specific numerical value of the angle at 3/4 height, and the assumptions regarding the observer's position and the flagpole's height are not explicitly stated.

oldfart1
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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
 
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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
 

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