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In fig.2 if $DE\parallel BC$, then what is $x$ equal to?

Apparently the answer is $10$, but how?

Question two:

If $$\sin{3x}=\cos{(x-6)}$$ where $3x$ and $x-6$ are both acute angles, find $x$.

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Hint: cos is $\pi/2$ shifted sine en.m.wikipedia.org/wiki/Sine –  Seyhmus Güngören Aug 29 '12 at 9:39
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1 Answer 1

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Ad question 2.

I'm assuming the angles are given in degrees, otherwise $3x, (x-6)\in \left(0,\frac{\pi}{2}\right)$ would provides a contradiction.
Since $\sin(90^\circ-(x-6))=\cos(x-6)$ the equality in question two could be simplified to $$\sin{3x}=\sin(90^\circ-(x-6)).$$ Sines of angles are equal if and only if the angles differ by $360^\circ k$ for some $k\in\mathbb{Z}$, or their sum equals an odd multiple of $180^\circ$. In other words $$\sin{\alpha}=\sin{\beta} \iff \alpha=\beta +360k \vee \alpha=180-\beta+360k,$$ so the expression is equivalent to $$(i)\space3x=90-x+6+360k$$or $$(ii)\space 3x = 180-(90-(x-6))+360k.$$ Both expressions can be easily simplified to $(i)\space x=24+90k$ or $(ii)\space x=42+180k$.
Now we have to find $k\in\mathbb{Z}$ which would ensure that $3x,(x-6)\in(0^\circ,90^\circ)$ [because the angles are said to be acute] i.e. $$x\in(0^\circ,30^\circ)$$ $$x\in(6^\circ,96^\circ)$$ must hold. The only $k$ for which $x=24+90k\in(6^\circ,30^\circ)$ is $k=0$. It is also clear that there is no $k\in \mathbb{Z}$ such that $x=42+180k\in(0^\circ,30^\circ)$, so apparently $(ii)$ provides no solution. Therefore the only one is $x=24^\circ.$

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Thanks, although i only understood a quarter of what you said(im only a beginner level, i dont even know what half that stuff means) but turns out my book was looking for the expression 3x = 90 - x + 6. YOUR A LIFESAVER! –  Aayush Agrawal Aug 29 '12 at 13:41
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