Question

to measure the height of the cloud cover at an airport, a worker shines a spotlight upward at an angle 75° from the horizontal. an observer d = 540 m away measures the angle of elevation to the spot of light to be 45°. find the height h of the cloud cover. (round your answer to the nearest meter.)

Explanation:

Step1: Set up equations using tangent function

Let the distance from the worker to the point directly below the cloud - cover be $x$ meters. The distance from the observer to the worker is $D = 540$ meters.
We know that $\tan45^{\circ}=\frac{h}{x + 540}$ and $\tan75^{\circ}=\frac{h}{x}$. Since $\tan45^{\circ}=1$, we have $h=x + 540$. And $\tan75^{\circ}=\tan(45^{\circ}+30^{\circ})=\frac{\tan45^{\circ}+\tan30^{\circ}}{1 - \tan45^{\circ}\tan30^{\circ}}=\frac{1+\frac{\sqrt{3}}{3}}{1 - 1\times\frac{\sqrt{3}}{3}}=\frac{3 + \sqrt{3}}{3-\sqrt{3}}=\frac{(3 + \sqrt{3})^2}{(3-\sqrt{3})(3 + \sqrt{3})}=\frac{9 + 6\sqrt{3}+3}{9 - 3}=\frac{12 + 6\sqrt{3}}{6}=2+\sqrt{3}$. Also, since $\tan75^{\circ}=\frac{h}{x}$, we have $x=\frac{h}{2 + \sqrt{3}}$.

Step2: Substitute $x$ into $h=x + 540$

Substitute $x=\frac{h}{2+\sqrt{3}}$ into $h=x + 540$, we get $h=\frac{h}{2+\sqrt{3}}+540$.
Rearrange the equation: $h-\frac{h}{2+\sqrt{3}}=540$. Factor out $h$: $h(1-\frac{1}{2+\sqrt{3}})=540$.
Simplify $1-\frac{1}{2+\sqrt{3}}=\frac{2+\sqrt{3}-1}{2+\sqrt{3}}=\frac{1+\sqrt{3}}{2+\sqrt{3}}$.
So $h=\frac{540(2+\sqrt{3})}{1+\sqrt{3}}$.
Rationalize the denominator: Multiply the numerator and denominator by $\sqrt{3}-1$.
$h=\frac{540(2+\sqrt{3})(\sqrt{3}-1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{540(2\sqrt{3}-2 + 3-\sqrt{3})}{3 - 1}=\frac{540(\sqrt{3}+1)}{2}=270(1 + \sqrt{3})\approx270\times(1 + 1.732)=270\times2.732 = 738$.

Answer:

$738$