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: Prove that $a\,+\,b\,+\,c=d$ enter image description here

share|improve this question
1  
Why do you say "buildings in the triangle" in all your geometry questions? I don't think it means whatever you think it means. –  Rahul Nov 4 '12 at 5:13
    
@Rahul Narain solutions such as none without using trigonometry:math.stackexchange.com/questions/223893/… –  felipeuni Nov 4 '12 at 5:30
    
That doesn't answer my question. But assuming you're Spanish, I guess when you say building you mean a construcción in geometry; this is called a geometrical construction in English. The word building is only used for a construcción in the architectural sense. –  Rahul Nov 4 '12 at 5:47
    
Sorry for the typo. I have just edited it. –  felipeuni Nov 4 '12 at 6:21

1 Answer 1

up vote 0 down vote accepted

enter image description here

Using sine law of triangle, $$\frac e{\sin(144^{\circ}-x)}=\frac g{\sin 36^{\circ}}$$ and $$\frac e{\sin(138^{\circ}-x)}=\frac {g+f}{\sin 42^{\circ}}$$

So, $$f=e\left(\frac{\sin 42^{\circ}}{\sin(138^{\circ}-x)}-\frac{\sin 36^{\circ}}{\sin(144^{\circ}-x)}\right)$$

$$=e\left(\frac{\sin 42^{\circ}\sin(144^{\circ}-x)-\sin 36^{\circ}\sin(138^{\circ}-x)}{\sin(138^{\circ}-x)\sin(144^{\circ}-x)}\right)$$

$2\sin 42^{\circ}\sin(144^{\circ}-x)$ $=\cos(102^{\circ}-x)-\cos({186^\circ}-x)$ using $2\sin A\sin B$ $=\cos(102^{\circ}-x)+\cos(6^\circ-x)$ as $\cos(180^{\circ}+y)=-\cos y$

$2\sin 36^{\circ}\sin(138^{\circ}-x)=\cos(102^{\circ}-x)-\cos({174^\circ}-x)$ $=\cos(102^{\circ}-x)+\cos(6^\circ+x)$ as $\cos({174^\circ}-x)=\cos\{180^\circ-(6^\circ+x)\}=-\cos(6^\circ+x)$

So, $$f=e\frac{\cos(6^\circ-x)-\cos(6^\circ+x)}{2\sin(138^{\circ}-x)\sin(144^{\circ}-x)}=\frac{e\sin x\sin 6^\circ }{\sin(138^{\circ}-x)\sin(144^{\circ}-x)}$$

For $f=a, x=12^\circ, a=\frac{e\sin 12^\circ\sin 6^\circ }{\sin126^{\circ}\sin132^{\circ}}$

For $f=b, x=60^\circ, b=\frac{e\sin 60^\circ\sin 6^\circ }{\sin78^{\circ}\sin84^{\circ}}$

For $f=c, x=96^\circ, c=\frac{e\sin 96^\circ\sin 6^\circ }{\sin42^{\circ}\sin48^{\circ}}$

$2\sin42^{\circ}\sin48^{\circ}=\cos 6^\circ -\cos 90^\circ=\cos 6^\circ$

and $\sin 96^\circ=\sin(90+6)^\circ=\cos 6^\circ$

So,$ c=2e\sin 6^\circ $

For $f=d, x=108^\circ, d=\frac{e\sin 108^\circ\sin 6^\circ }{\sin30^{\circ}\sin36^{\circ}}=\frac{2e\sin 72^\circ\sin 6^\circ }{\sin36^{\circ}}=4e\cos 36^\circ \sin 6^\circ$

share|improve this answer

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.