Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x$.

share|cite|improve this question
Hint: cos is $\pi/2$ shifted sine – Seyhmus Güngören Aug 29 '12 at 9:39
up vote 1 down vote accepted

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 $x=24^\circ.$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.