Given Quadrilateral ABCD in such that $AB<BC<CD$ creating increasing arithmetic progression with sum of $27$ cm.

$\measuredangle BCD=60^{0}$. the diagonal $BD=\sqrt{133}$ cm, and it divided $\measuredangle ABC$ such that:

$\measuredangle CBD=2\measuredangle DBA $ . compute the length of diagonal $AC$.

This is a bit strange because computing by law of sines or law of cosines will give different number of solutions and i can't really prove that one of the solutions is disqualified.

If we look at triangle $BCD$: $\frac{\sqrt{133}}{sin60^{0}}=\frac{9}{sin\measuredangle BDC}$ so $\measuredangle BDC=42.5^{0}$ and $\measuredangle DBC=77.48^{0}$

But if i compute: $\frac{\sqrt{133}}{sin60^{0}}=\frac{13}{sin\measuredangle DBC}$ so $\measuredangle DBC=77.48^{0}$ or $102.51^{0}$

  • $\begingroup$ It will help on this end if you show what you found, so someone here can tell you which one of your solutions worked right or went wrong... $\endgroup$ May 1, 2013 at 19:59
  • $\begingroup$ @RecklessReckoner - did it right now. $\endgroup$
    – getter
    May 1, 2013 at 20:16
  • $\begingroup$ i would appreciate if someone can explain why i should think from the start to rule out one of the solution if i use law of sine such that angle $DBC=77.48^{0} or 102.51^{0}$? it fits the problem and i don't think that i need to compute again using law of sine differently just to rule out the other angle. in addition: is there any restriction of using law of cosine? maybe the 2 answers would fit aand then cosine wouldn't get me there. $\endgroup$
    – getter
    May 1, 2013 at 21:02
  • $\begingroup$ You say the perimeter is 27? But in your 2nd solution you seem to identify length(CD) = 13. Is that right? Where did that 13 come from? $\endgroup$
    – Mark Ping
    May 1, 2013 at 21:26
  • $\begingroup$ The 13 comes from applying the Law of Cosines and the arithmetic sequence requirement that CD have length $9 + d$ to get the diagonal BD to have length $\sqrt{133}$. $\endgroup$ May 1, 2013 at 21:35

1 Answer 1


Let $AB=x$, $BC=x+y$, $CD=x+2y$. The sum is $27$, so $x+y=9$.

Now use the Cosine Law. We have $CD=9+y$, and therefore $$133=(9+y)^2+81-(2)(9)(9+y)(1/2).$$ This simplifies to $y^2+18y-52=(y+13)(y-4)=0$, so $y=4$. Thus our arithmetic progression is $5,9,13$.

Now we can use the Cosine Law again to find the cosine of $\angle CBD$, and then trigonometric identities to find the cosine of $\angle ABC$, and then the Cosine Law to find $AC$. Although exact expressions can be found, they are not particularly attractive. The angle $ABC$ turns out to be approximately $116.22$ degrees.

Remark: Using the Sine Law is less efficient: the cosine identifies the angle of a triangle uniquely, but the sine does not. So using the Sine Law to compute $\angle DBC$ leads to extra work. There is only one possible answer, since the sides determine the angles.

We can rule out the possibility that $\angle DBC$ is obtuse in various ways. For example, we can compute the sine of $\angle CDB$, the smallest angle, by using the Sine Law. There will be no ambiguity, the angle turns out to be about $42.52^\circ$. Now the remaining angle $DBC$ gets uniquely identified: it is $\approx 180-(60+42.52)$.

  • $\begingroup$ thanks but i didn't get why my other solution is wrong. $\endgroup$
    – getter
    May 1, 2013 at 20:24
  • $\begingroup$ that's one way. how can you explain that if i compute by the other way i get angle DBC $77.48^{0}$ or $102.51^{0}$? in that way i can't see what's wrong. it's strange that i need to solve this question with 2 different ways to get the right answer. $\endgroup$
    – getter
    May 1, 2013 at 20:50
  • $\begingroup$ the question is: why should i thinks that it would be a problem (extra work) if it fits by the sum of angles in a triangle? $\endgroup$
    – getter
    May 1, 2013 at 21:05
  • $\begingroup$ no, im asking why if 2 options fits the sum of angles in a triangle is should consider checking those solutions with another approach? $\endgroup$
    – getter
    May 1, 2013 at 21:23
  • $\begingroup$ There is only one option here that makes the sum of the angles $180^\circ$, for choosing the obtuse angle will violate the Sine Law for the third angle $CDB$. The system is objecting to the length of the comment string. I will delete most of mine, suggest you do the same. $\endgroup$ May 1, 2013 at 21:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .