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Any idea about this problem:

As shown in the figure: Prove that $X=30.$

enter image description here

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First of all, what's important to know when dealing with triangle problems like this is: Is trigonometry allowed? Because that makes this problem and many others like it a whole lot easier. –  Arthur Jan 5 '13 at 0:16
    
@Arthur The original two tags posted by the OP included (trigonometry) and (triangle), so I suspect it's safe to assume one can use trigonometry. –  amWhy Jan 5 '13 at 0:22
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felipeuni: It would be helpful if you would show some work, and asked only about a more specific roadblock you've encountered. Surely, there are properties and values (lengths, angles) that you can deduce immediately from the given information, and it would help if you'd start by cataloging that, and then outline what you need to know to determine angle X...then work toward finding that information. That is not to say that the problem is trivial. But we don't want to tell you what you already know. –  amWhy Jan 5 '13 at 0:24

2 Answers 2

up vote 3 down vote accepted

If $AH=1$ then $HQ=\tan(27^\circ),HP=\tan(63^\circ),AC=1+\tan(66^\circ)\tan(42^\circ)$

Then $\vec{CQ}=(1-AC,HQ),\vec{CP}=(1-AC,HP)$

And finally $\angle \vec{CQ},\vec{CP} = \arccos \frac{\vec{CQ}\cdot\vec{CP}}{|\vec{CQ}||\vec{CP}|}=30^\circ$

The tricky part is to simplify this $\frac{\tan ^2(42 {}^{\circ}) \cot ^2(24 {}^{\circ})+1}{\sqrt{\left(\cot ^2(27 {}^{\circ})+\tan ^2(42 {}^{\circ}) \cot ^2(24 {}^{\circ})\right) \left(\tan ^2(27 {}^{\circ})+\tan ^2(42 {}^{\circ}) \cot ^2(24 {}^{\circ})\right)}}$ to be $\cos(30^\circ)$.

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Isn't $HQ = \tan 27^\circ$? –  Calvin Lin Jan 5 '13 at 0:45
    
The 'tricky' part isn't obvious why it should hold. Arguably, that's where the work is. –  Calvin Lin Jan 5 '13 at 0:55
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Are you going to serve the OP's other "order"? math.stackexchange.com/questions/235467/…, or perhaps this "order": math.stackexchange.com/questions/228700/…. The OP has a habit of assigning us problems to solve for him/her. –  amWhy Jan 5 '13 at 0:56
    
@amWhy Nah, I'm done here. I was just a bit bored. –  swish Jan 5 '13 at 0:57
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swish I didn't mean to be sarcastic (or perhaps shouldn't have been even though I was). I just get a little annoyed when an asker repeatedly, despite requests for more information, or requests to show some work, posts questions to be solved here. –  amWhy Jan 5 '13 at 1:00

Hints:

0) Did you happen to notice the big triangle is isosceles...?

1) Second acute angle in $\,\Delta AHQ\,$ is $\,63^\circ\,$

2) $\,\angle AQP=117^\circ\,$

3) $\,\angle APQ= 27^\circ\,...$

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I'm made those observations too, but am unable to proceed. Can you explain how to continue? –  Calvin Lin Jan 5 '13 at 0:08
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@DonAntonio I did. It seems to me that you just made those observations quickly and underestimated the question. The observations you made are quite trivial and OP probably made them too. It's a good start, that I agree with. –  ZafarS Jan 5 '13 at 0:14
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So you know, @ZafarS, that the observations are so trivial that the OP ""probably"" made them...but we haven't heard from him/her, yet. Interesting. I'd rather wait for the OP to write about her/his thoughts, ideas and, who knows, perhaps even something he could have received frm my answer, and then I will give my opinion. The downvote doesn't seem a good and fair idea, and what goes around comes around. –  DonAntonio Jan 5 '13 at 0:17
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What goes around comes around? Extremely childish, sir. I'll remove the downvote when you update the answer, or when OP makes it clear your hints were sufficient. At this juncture, this isn't a good answer. –  ZafarS Jan 5 '13 at 0:19
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@DonAntonio Going the way you're hinting at, you'd hit a dead end around hint number 4, and you'd most likely be stuck (I know I am). Very stuck, if you're trying to look for similarily trivial continuations. –  Arthur Jan 5 '13 at 0:21

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