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A pole fixed on the ground is leaning away from the vertical. When the Sun is directly overhead, the length of its shadow is 7.5m. An observer standing 30m away from the base of the pole in the direction of the shadow, measures the angle of elevation of the top of the pole to be 30o. Find the length of the pole. Ignore the height of the observer,

(A) 7.5m (B) 15m (C) 22.5m (D) 30m

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Welcome to math.SE: since you are a new user, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many find the use of imperative ("Find") to be rude when asking for help; please consider rewriting your post. –  Zev Chonoles Apr 24 '12 at 10:24
    
Please rephrase as a question. Also, it is helpful to explain where you are stuck. Have you drawn a diagram? –  Tpofofn Apr 24 '12 at 10:37
    
@ Zev Chonoles NO this is not a homework question. –  Abhishek Pant Apr 24 '12 at 13:48
2  
Why do you word your question as a multiple-choice problem if it is not homework (or the moral equivalent thereof)? –  Henning Makholm Apr 24 '12 at 14:39
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1 Answer

up vote 2 down vote accepted

An "angle of elevation" of $300$ degrees is very non-standard. In the ordinary meaning of the term, the angle of elevation is always less than $90$ degrees. I will assume that $300$ is a typo for $30^\circ$. There is some evidence for that: with an angle of elevation of $30^\circ$, the numbers turn out very nice. We get the good old $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangle. and one of the given options is correct. With other angles of elevation, the calculation can be done in a similar way, the numbers are just uglier.

Draw a diagram. Label the top of the pole $T$, the bottom of the pole $B$, and the location of the observer $O$. Drop a perpendicular from the top $T$ of the pole to the line $BT$. Suppose this meets the line segment $BT$ at $S$, the tip of the shadow of the pole.

Look at $\triangle TBS$, and let $\angle TBS=\theta$. Let $p$ be the length of the pole. Then $$\cos\theta=\frac{7.5}{p},\quad\text{and therefore}\quad7.5=p\cos\theta. \tag{$1$}$$ We are told that $BS=7.5$. Thus $SO=30-7.5=22.5$. Note that $ST=p\sin\theta$. We have $$\frac{ST}{SO}=\frac{p\sin\theta}{22.5}=\tan(30^\circ)=\frac{1}{\sqrt{3}}.$$ It follows that $$p\sin\theta=\frac{22.5}{\sqrt{3}}=7.5\sqrt{3}.\tag{$2$}$$ Now $$(p\cos\theta)^2+(p\sin\theta)^2=p^2(\cos^2\theta+\sin^2\theta)=p^2.$$ We conclude from $(1)$ and $(2)$ that $$(7.5)^2+ (7.5\sqrt{3})^2=p^2.$$ This simplifies to $(7.5)^2(4)=p^2$, from which it follows that $p=(7.5)(2)=15$.

Remarks: $1.$ It would be more standard, and easier in this case, to observe that $SO=22.5$ and since $\frac{ST}{SO}=\tan(30^\circ)$, we have $ST=(22.5)/\sqrt{3}=7.5\sqrt{3}$. Then by the Pythagorean Theorem, $(BT)^2=(ST)^2+(BT)^2=(7.5\sqrt{3})^2+(7.5)^2$, and we are finished.

$2.$ I did not pay attention to the choices provided in this multiple choice question. However, the answers provided here are quite far from each other. It might be best test-taking strategy to rule out the incorrect answers by making a few sketches. In this case, that is far faster than the method that I used.

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thanks very much. yes it's a typo mistake. –  Abhishek Pant Apr 24 '12 at 13:45
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