Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Geometry: Auxiliary Lines

As shown in the figure: enter image description here

share|improve this question
1  
The question seems so simple, but I can't do it. –  ᴊ ᴀ s ᴏ ɴ Nov 4 '12 at 6:51
    
Almost the same as your previous question: math.stackexchange.com/questions/228534/… –  Joel Reyes Noche Nov 4 '12 at 11:05

2 Answers 2

up vote 1 down vote accepted

This is not the desired method, but it does show that $50^\circ$ is the result.

Call the fourth point D, and suppose A is at $(0,0)$ and C is at $(1,0)$. Then B is at $(x_B,y_B)=\left(\frac{\tan(70^\circ)}{\tan(58^\circ)+\tan(70^\circ)},\frac{\tan(58^\circ)\tan(70^\circ)}{\tan(58^\circ)+\tan(70^\circ)}\right) \approx (0.631921861,1.011286374)$ while D is at $(x_D,y_D)=\left(\frac{\tan(62^\circ)}{\tan(12^\circ)+\tan(62^\circ)},\frac{\tan(12^\circ)\tan(62^\circ)}{\tan(12^\circ)+\tan(62^\circ)}\right) \approx (0.898457801,0.190973101)$.

Then the length of AD is $a=\sqrt{x_D^2+y_D^2} \approx 0.918529883$ and of BD is $b = \sqrt{(x_D-x_B)^2+(y_D-y_C)^2} \approx 0.862528419$. Using the sine rule, the angle ABD is $$\sin^{-1}\left(\frac{a}{b}\sin(46^\circ)\right) = 50^\circ$$ at least to the precision of my calculations.

For what it is worth, an accurate diagram looks like

enter image description here

share|improve this answer

Let central point is $O$.

$\angle OBC = 180^\circ - (46^\circ+12^\circ) - (62^\circ+8^\circ) - X = 52^\circ-X$.

Using the sine formula, we have: $$\frac{AO}{\sin 62^\circ} = \frac{CO}{\sin 12^\circ},$$ $$\frac{AO}{\sin X} = \frac{BO}{\sin 46^\circ},$$ $$\frac{CO}{\sin (52^\circ-X)} = \frac{BO}{\sin 8^\circ},$$ so, $$\sin 12^\circ \cdot \sin 8^\circ \cdot \sin X = \sin 62^\circ \cdot \sin 46^\circ \cdot \sin (52^\circ-X).$$

(Here we can find that $X=50^\circ$. Uniqueness of solution is discussed in comments).


In fact we need to prove that $$\sin 8^\circ \cdot \sin 12^\circ \cdot \sin 50^\circ =^{???} \sin 2^\circ \cdot \sin 46^\circ \cdot \sin 62^\circ.$$ Using formulas $$2\cdot \sin\alpha \cdot \sin\beta = \cos(\alpha-\beta) - \cos(\alpha+\beta),$$ $$2\cdot \sin\alpha \cdot \cos\beta = \sin(\alpha-\beta) + \sin(\alpha+\beta),$$ $$\sin(180^\circ-\varphi) = \sin(\varphi), \quad \cos(180^\circ-\varphi) = -\cos(\varphi),$$

we have $$\sin 8^\circ \cdot (\cos 38^\circ - \cos 62^\circ) =^{???} \sin 2^\circ \cdot (\cos 16^\circ + \cos 72^\circ),$$ $$-\sin 30^\circ + \sin 46^\circ + \sin 54^\circ - \sin 70^\circ =^{???} -\sin 14^\circ + \sin 18^\circ - \sin 70^\circ + \sin 74^\circ,$$ $$-\sin 30^\circ + \sin 46^\circ + \sin 54^\circ + \sin 14^\circ - \sin 18^\circ - \sin 74^\circ =^{???} 0,$$ $$(\sin 14^\circ + \sin 46^\circ - \sin 74^\circ) + ( - \sin 18^\circ + \sin 54^\circ -\frac{1}{2} )=^{???} 0,$$ $$(\sin 14^\circ + \sin 134^\circ + \sin 254^\circ) + \frac{1}{2}( - \sin 18^\circ + \sin 54^\circ +\sin 126^\circ + \sin 198^\circ + \sin270^\circ )=^{???} 0,$$ it is equality, because left side is equal to $$\mathrm{Im} \left( p(\omega_3^0 + \omega_3^1 + \omega_3^2) + \frac{q}{2}( \omega_5^0 + \omega_5^1 +\omega_5^2 + \omega_5^3 + \omega_5^4) \right)=$$ $$\mathrm{Im} \left( p \cdot \frac{\omega_3^3-1}{\omega_3-1} + \frac{q}{2}\cdot \frac{\omega_5^5-1}{\omega_5-1} \right)= \mathrm{Im} \left( p \cdot \frac{1-1}{\omega_3-1} + \frac{q}{2}\cdot \frac{1-1}{\omega_5-1} \right)= \mathrm{Im} \left( p \cdot 0 + \frac{q}{2} \cdot 0 \right)= 0,$$ where $p = \exp(i\pi 7/90)$, $q = \exp(-i\pi /10)$, $w_{...}$ $-$ roots of unity:

$w_3 = \exp(i 2 \pi / 3) = \cos 2\pi/3 + i \sin 2\pi/3 = \cos 120^\circ + i \sin 120^\circ$,

$w_5 = \exp(i 2 \pi / 5) = \cos 2\pi/5 + i \sin 2\pi/5 = \cos 72^\circ + i \sin 72^\circ$.

share|improve this answer
2  
How do you find that $X=50$? –  wj32 Nov 4 '12 at 10:55
    
Here I need to prove that $\sin 12^\circ \cdot \sin 8^\circ \cdot \sin 50^\circ = \sin 62^\circ \cdot \sin 46^\circ \cdot \sin 2^\circ$. I will think about it... Not trivial... –  Oleg567 Nov 5 '12 at 16:16
    
@wj32: since (for postive angles less than rightangles) $\sin(x)$ is a function increasing monotonically from $0$ to $1$, there is a unique solution. And $50^\circ$ works. –  Henry Nov 7 '12 at 11:22
    
@Henry: Well that's part of what I'm asking - how do you prove that $50^\circ$ works? –  wj32 Nov 7 '12 at 11:24
    
@wj32: I updated proof of $50^\circ$ (look above) now. –  Oleg567 Nov 7 '12 at 11:40

Your Answer

 
discard

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.