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 $x$.
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 $x$.
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 $x=24^\circ.$