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... $\endgroup$ – colormegone May 1 '13 at 19:59
  • $\begingroup$ @RecklessReckoner - did it right now. $\endgroup$ – getter May 1 '13 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 '13 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 '13 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 $\sqrt{133}$. $\endgroup$ – colormegone May 1 '13 at 21:35

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 '13 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 '13 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 '13 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 '13 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. $\endgroup$ – André Nicolas May 1 '13 at 21:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.