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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 $D = 500 m$ 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.)

So what I did was take the tangent of both angles:

$tan45°=\frac{h}{500-D}$

$tan75°=\frac{h}{D}$

What do I do next? I tried using substitution but I got $h=~105.7$ which is wrong.

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    $\begingroup$ I thought D was 500? In any case, you should know the tangent of 45 degrees. $\endgroup$ – lulu Jul 23 '15 at 2:58
  • $\begingroup$ The observer is actually $D=500$m away *from the spotlight*, not from the intersection of $h$ with the ground. $\endgroup$ – coldnumber Jul 23 '15 at 3:07
  • $\begingroup$ Wait so doing $h/500-D$ and $h/D$ is wrong? What would I do instead? $\endgroup$ – TheNewGuy Jul 23 '15 at 3:14
  • $\begingroup$ Just replace $D$ in your equation with an unknown distance $x$, and then you can get $x$ in terms of $h$. (See my answer.) $\endgroup$ – coldnumber Jul 23 '15 at 3:15
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The observer is 500 m away from the spotlight, but you can let $x$ be the distance from the observer to the intersection of the segment of height $h$ with the floor.

Then you get the equations $\tan{45^\circ}=\dfrac{h}{x}$ and $\tan{75^\circ}=\dfrac{h}{500-x}$.

Since $\tan{45^\circ}=1$, the first equation tells you that $h=x$, and then you can plug $h$ into the second equation and solve for $h$.


EDIT: Solving the second equation for $h$

Above we found that $x=h$, so the second equation becomes \begin{align} \tan{75^\circ}=\frac{h}{500-h} &\iff h=(500-h)\tan{75^\circ} \\ &\iff h=500\tan{75^\circ}-h\tan{75^\circ} \\ &\iff h+h\tan{75^\circ}=500\tan{75^\circ} \\ &\iff h(1+\tan{75^\circ})=500\tan{75^\circ}\\ &\iff h=\frac{500\tan{75^\circ}}{1+\tan{75^\circ}} \end{align}

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  • $\begingroup$ I got 105.7, is that what you got? $\endgroup$ – TheNewGuy Jul 23 '15 at 3:23
  • $\begingroup$ Nope. I got $\tan 75^\circ = \frac{h}{500-h} \implies h=\frac{500\tan 75^\circ}{1+\tan 75^\circ}$, which is approximately 394.337. $\endgroup$ – coldnumber Jul 23 '15 at 3:25
  • $\begingroup$ You're answer is correct but I don't understand the last part. I got the two equations and I understand how you got $h = x$, but I don't understand how you plugged it in and got $h=\frac{500tan75}{1+tan75}$ $\endgroup$ – TheNewGuy Jul 23 '15 at 3:38
  • $\begingroup$ OK. let me add extra steps to my answer. $\endgroup$ – coldnumber Jul 23 '15 at 3:39
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Hint: $$ h\cot 45^\circ + h\cot75^\circ = D $$

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  • $\begingroup$ I factored out the h and got $h(tan45+tan75)=500$. Then I divided $(tan45+tan75)$ by both sides and got $h= ~ 105.7$ $\endgroup$ – TheNewGuy Jul 23 '15 at 3:10
  • $\begingroup$ I think your equation should be $D=\frac{h}{\tan 45^\circ}+\frac{h}{\tan 75^\circ}$ $\endgroup$ – coldnumber Jul 23 '15 at 3:17
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    $\begingroup$ @coldnumber, oh, really. Thanks $\endgroup$ – Michael Galuza Jul 23 '15 at 3:19
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    $\begingroup$ @HarishChandraRajpoot, it is answer. If you prefer explain to OP's why $1+1=2$ please go ahead. Comment to downarrow: "The answer is not useful". This hint (actually) is useful. Think about it $\endgroup$ – Michael Galuza Jul 23 '15 at 3:27
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    $\begingroup$ A sufficient hint IMHO (+1) $\endgroup$ – Jyrki Lahtonen Jul 23 '15 at 6:50
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Take cloud as C,Take foot of perpendicular as P,Take observer as O,Take worker as W $tan 45=\frac{CP}{OP}=\frac{h}{OP}$
$1=\frac{h}{OP}$
$OP=h$
$tan 75=\frac{h}{PW}$
$tan (45+30)=\frac{h}{500-OP}$
$\frac{tan 45+tan30}{1-tan45 tan 30}=\frac{h}{500-h}$
$\frac{1+\frac{1}{\sqrt3}}{1-\frac{1}{\sqrt3}}=\frac{h}{500-h}$
$\frac{\sqrt3+1}{\sqrt3-1}=\frac{h}{500-h}$
$\frac{\sqrt3+1}{\sqrt3-1}\times\frac{\sqrt3+1}{\sqrt3+1}=\frac{h}{500-h}$
$2+\sqrt3=\frac{h}{500-h}$
Put $\sqrt3=1.732$ and solve

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Let $\tan(45^{\circ})=h/x_{1}$ and $\tan(75^{\circ})=h/x_{2}$ such that $D=x_{1}+x_{2}$. Since $$\tan(75^{\circ})=\frac{h}{x_{2}}=\frac{h}{D-x_{1}}\implies x_{1}=D-\frac{h}{\tan(75^{\circ})}=\frac{D\tan(75^{\circ})-h}{\tan(75^{\circ})},$$ then you have $$\tan(45^{\circ})=\frac{h}{x_{1}}=\frac{h\tan(75^{\circ})}{D\tan(75^{\circ})-h}\iff h=\frac{\tan(45^{\circ})D\tan(75^{\circ})}{\tan(45^{\circ})+\tan(75^{\circ})}.$$ Hence

$$h=\frac{\tan(45^{\circ})\cdot 500\cdot \tan(75^{\circ})}{\tan(45^{\circ})+\tan(75^{\circ})}\approx 394.33757\quad m$$

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