# Thread: Flagpole and Building

1. ## Flagpole and Building

A 50-ft flagpole stands on the top of a 25-ft building. How far from the base of the building should a person stand if the flagpole and the building are to subtend equal angles at her eye, which is 5 ft above the ground?

2. ## Re: Flagpole and Building

Did you make a sketch?

$\dfrac{20}{x} = \tan{\theta}$

$\dfrac{70}{x} = \tan(2\theta)$

solve the system for $x$ ...

3. ## Re: Flagpole and Building

when I solve the system, I get tan x = square root of 3/7

x = 20 / square root of 3/7 = 30

but the book answer is 40 4. ## Re: Flagpole and Building

My mistake ... I misinterpreted the set up. Let me work on it again.

5. ## Re: Flagpole and Building

$\tan{\alpha} = \dfrac{5}{x}$

$\tan{\beta} = \dfrac{20}{x}$

$\tan(\theta+\beta) = \dfrac{70}{x}$

$\alpha + \beta = \theta$

a bit more complex of a system, but solvable ...

6. ## Re: Flagpole and Building

wOW. Very difficult to solve it algebraically

7. ## Re: Flagpole and Building

There is a simpler solution possible using the angle bisector theorem ...

The segment from eyeball to top of building cuts the opposite side in a ratio of 25:50, or 1:2

Adjacent sides of large triangle (eyeball to base of building & eyeball to top of flag) will also be in a 1:2 ratio.

Pythagoras ...

$x^2+5^2=a^2$

$x^2+70^2=(2a)^2 \implies x^2+70^2=4(x^2+5^2) \implies 4800=3x^2 \implies x = 40$

8. ## Re: Flagpole and Building

wOW, it is very easy to solve using the theorem.
Does the angle bisector always give adjacent sides the ratio 1:2?

9. ## Re: Flagpole and Building Originally Posted by joshuaa wOW, it is very easy to solve using the theorem.
Does the angle bisector always give adjacent sides the ratio 1:2?
see link ... click on the question mark (?)

angle-bisector-theorem

10. ## Re: Flagpole and Building

wOW, that's fantastic and very helpful. thanks a Lot.

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