Question

two observers are 600 ft apart on opposite sides of a flagpole. the angles of elevation from the observers to the top of the pole are 17° and 20°. find the height of the flagpole. the flagpole is ft high. (round to the nearest tenth as needed.)

Explanation:

Step1: Set up tangent - based equations

Let the height of the flag - pole be $h$ feet. Let the distance of the observer with an angle of elevation of $17^{\circ}$ from the base of the flag - pole be $x$ feet. Then the distance of the other observer (with an angle of elevation of $20^{\circ}$) from the base of the flag - pole is $(600 - x)$ feet.
We know that $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. So, $\tan17^{\circ}=\frac{h}{x}$, which implies $x = \frac{h}{\tan17^{\circ}}$, and $\tan20^{\circ}=\frac{h}{600 - x}$, which implies $600 - x=\frac{h}{\tan20^{\circ}}$.

Step2: Substitute $x$ into the second equation

Substitute $x=\frac{h}{\tan17^{\circ}}$ into $600 - x=\frac{h}{\tan20^{\circ}}$. We get $600-\frac{h}{\tan17^{\circ}}=\frac{h}{\tan20^{\circ}}$.

Step3: Rearrange the equation to solve for $h$

First, get a common denominator. Multiply through by $\tan17^{\circ}\tan20^{\circ}$:
$600\tan17^{\circ}\tan20^{\circ}-h\tan20^{\circ}=h\tan17^{\circ}$.
Then, move the terms with $h$ to one side:
$600\tan17^{\circ}\tan20^{\circ}=h\tan17^{\circ}+h\tan20^{\circ}$.
Factor out $h$: $h=\frac{600\tan17^{\circ}\tan20^{\circ}}{\tan17^{\circ}+\tan20^{\circ}}$.
We know that $\tan17^{\circ}\approx0.3057$ and $\tan20^{\circ}\approx0.3640$.
$h=\frac{600\times0.3057\times0.3640}{0.3057 + 0.3640}$.
$h=\frac{600\times0.1113748}{0.6697}$.
$h=\frac{66.82488}{0.6697}\approx99.8$.

Answer:

$99.8$