# In quadilateral $ABCD$, $AB=16\sqrt{2}$ cm, $CD=10$ cm, $DA=8.5$ cm, $\angle D = 120^\circ$ and $\angle ACB = 45^\circ$. How to find $\angle ABC$?

In quadilateral $ABCD$ (usual clockwise or anticlockwise naming), $AB=16\sqrt{2}$ cm, $CD=10$ cm, $DA=8.5$ cm, $\angle D = 120^\circ$ and $\angle ACB = 45^\circ$. How to find $\angle ABC$?

As stated in one of the answer, the obvious approach, utilizing the law of cosines and sines gives a very ugly form for a problem that is intended for pencil-paper calculation. I was wondering if there is any alternative approach to avoid doing the messy parts?

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If a problem is intended for a paper-pencil calculation, then the calculations should be exact. If you ask a mathematician to solve this problem, he will definitely give you the exact value of the angle as $\arcsin( something)$. That may be close to 30 degrees, but it is not 30 degrees. –  Beni Bogosel Oct 26 '11 at 20:51
@Beni Bogosel: Quantitative aptitude is a different flavor of mathematics,it's not about generalize but more often particularize,use of various tricks,approximations and one's only goal is to choose the correct options and nothing else. –  Quixotic Oct 26 '11 at 21:10

Using the Law of Cosines, I get that $|AC|^2=8.5^2+10^2+85=257.25$ since $\cos(ADC)=-\frac{1}{2}$. Next, $\sin^2(ACB)=\frac{1}{2}$ and $|AB|^2=512$. Law of Sines says that $$\frac{\sin^2(ACB)}{|AB|^2}=\frac{\sin^2(ABC)}{|AC|^2}$$ Therefore, $$\sin^2(ABC)=\frac{1}{2}\frac{257.25}{512}\approx\frac{1}{4}$$ Thus, $ABC$ must be about $30^\circ$. The hardest thing to do was square $8.5$.

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Any reason from the downvoter? –  robjohn Oct 26 '11 at 22:38

My guess is:

• find $AC$ applying cosine law in triangle $ADC$;

• apply sine theorem in triangle $ABC$: $$\frac{AB}{\sin \angle ACB}=\frac{AC}{\sin\angle ABC}$$

From the last relation you should be able to find $\sin B$ and then $B$. I think that your side lengths are wrong, because it gets very messy.

You can use the calculator from over here to do the calculations: http://www.handymath.com/cgi-bin/irregangle8.cgi

The angle is about $30$ degrees, but not exactly.

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Ok, now, if it is the question of the day and you want to participate, why do you ask here for help? You should ask when the deadline is passed... –  Beni Bogosel Oct 25 '11 at 20:53
:It's not giving any money or prize,and also the solution they give are silly and not rigorous,but here I have the opportunity to disscuss and learn new cool things ;)You may argue that I could do the same thing tomorrow,but I don't know if I would be here tomorrow... –  Quixotic Oct 25 '11 at 20:56
Have you at least tried something? At least do the calculations which arise using the formulas I provided, to see what I mean. –  Beni Bogosel Oct 25 '11 at 20:58
Yes,I noticed that things is getting messy.But may be there is another approach to get rid of messy things. –  Quixotic Oct 25 '11 at 21:11
I used mathematica,$AC \approx 16.039$ which in terms gives $\sin x^\circ = 0.501219 \Rightarrow$,x is close to $30^\circ$. –  Quixotic Oct 25 '11 at 21:19