Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


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?

share|cite|improve this question
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
up vote 3 down vote accepted

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 $8.5$.

share|cite|improve this answer
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:

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

share|cite|improve this answer
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

Your Answer


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.