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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 $\angle ABC$?

Problem source.

ADDED:

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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    @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.2011-10-26

2 Answers 2

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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?2011-10-26
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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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    I used mathematica,$AC \approx 16.039$ which in terms gives $\sin x^\circ = 0.501219 \Rightarrow $,x is close to $30^\circ$.2011-10-25