Question

jason is standing 100 feet from the base of a flag pole. he is looking up to the top of the pole at an angle of 30°. if his eye level is 5.5 feet, how tall is the flagpole?

Explanation:

Step1: Define height segment above eye level

Let $h$ = height from eye level to top of pole. Use tangent of $30^\circ$:
$\tan(30^\circ) = \frac{h}{100}$

Step2: Solve for $h$

Rearrange to isolate $h$, substitute $\tan(30^\circ)=\frac{1}{\sqrt{3}}$:
$h = 100 \times \tan(30^\circ) = 100 \times \frac{1}{\sqrt{3}} = \frac{100\sqrt{3}}{3} \approx 57.74$

Step3: Total pole height calculation

Add eye level height to $h$:
$\text{Total Height} = \frac{100\sqrt{3}}{3} + 5.5$

Answer:

The height of the flagpole is $\frac{100\sqrt{3}}{3} + 5.5$ feet, or approximately 63.24 feet.