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March 21st, 2019, 05:44 PM  #1 
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  How to find the smallest length in a triangle when the angles aren't known?
I am confused on how to find the answer for this problem. So far what I believe would apply is the triangle inequality but I'm not sure on how to use it. The figure $ABC$ is a triangle so as $BDC$. It is known that the length of $AB = 8$ inches and $\angle BAD = 2\,\angle BCD$ and $\angle\,DBC = 3 \angle BCD$. Find the smallest whole number that $BD$ can attain. The alternatives given are: $\begin{array}{ll} 1.&7\,\textrm{inches}\\ 2.&4\,\textrm{inches}\\ 3.&8\,\textrm{inches}\\ 4.&5\,\textrm{inches}\\ 5.&6\,\textrm{inches}\\ \end{array}$ So far the only thing I could come up with was to "guess" it might be $4$ since, $\angle BDA = 4\omega$ could it be that since $BD$ is opposing $2\omega$ hence that side might be $4$ inches. But that's how far I went. Needless to say that I'm unsure if that's a right way to go. 
March 21st, 2019, 07:11 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,941 Thanks: 1545 
Using the Law of sines ... $\dfrac{8}{\sin(4\omega)} = \dfrac{BD}{\sin(2\omega)}$ $\dfrac{8}{2\sin(2\omega)\cos(2\omega)} = \dfrac{BD}{\sin(2\omega)}$ $\dfrac{4}{\cos(2\omega)} = BD$ $0 < \cos(2\omega) < 1 \implies BD > 4$ I'd say minimum of 5" 
March 22nd, 2019, 09:43 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,580 Thanks: 1038 
BD is shortest when perpendicular to AC. So triangle ABD is right/isosceles: do the work and you'll end up with ~5.656.... thus 5 as per Skeeter. 
March 22nd, 2019, 06:17 PM  #4  
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  More help please Quote:
$\frac{4}{\cos 2\omega}= BD$ How does this equality becomes into an inequality? The range you given as: $0 < \cos 2\omega < 1$ How does it imply this? $BD > 4$ I don't know how to translate the words "minimum value" from those equations: So far I think that if they ask the smallest value for $BD$ then does it mean $\cos 2\omega$ should be close to $1$ so as when it is inserted in the equation it will not increase the factor of $4$ in the numerator. But from there I'm still lost. What am I missing?. Can you explain me more, please?  
March 22nd, 2019, 06:23 PM  #5 
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  I wonder how did you obtained 5.656? But yes it makes sense that it will be the smallest when perpendicular. It took me a while to find out where the isosceles triangle you were referring to is, but I found it. But from there I'm still stuck. Maybe can you offer an additional hint? Last edited by skipjack; March 23rd, 2019 at 07:46 AM. 
March 23rd, 2019, 05:23 AM  #6  
Math Team Joined: Jul 2011 From: Texas Posts: 2,941 Thanks: 1545  Quote:
$0 < \cos(2\omega) < 1$ substitute $\dfrac{4}{BD}$ for $\cos(2\omega)$ in the above inequality $0 < \dfrac{4}{BD} < 1$ multiply every term in the inequality by the positive value $BD$ ... $0 < 4 < BD$  
March 23rd, 2019, 06:30 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,580 Thanks: 1038  As I told you, I ASSUMED that the diagram can be rearranged, with BD perpendicular to AC. This results in w = 22.5 degrees. Triangle ABD becomes a right triangle with hypotenuse AB = 8, and AD = BD, thus isosceles; AD = BD = ~5.6568 Looks like that was NOT the intent of your problem. WHAT really does "smallest whole number that BD can attain" mean? Is that a translation from another language? 
March 23rd, 2019, 07:53 AM  #8 
Math Team Joined: Jul 2011 From: Texas Posts: 2,941 Thanks: 1545 
Working it out, $BD =5 \implies \omega \approx 18.4^\circ$ therefore, $\angle{A} \approx 36.9^\circ \text{ and }\angle{BDA} = 4\omega \approx 73.7^\circ \implies \angle{ABD} \approx 69.4^\circ$ Further, $\angle{ABC} = 124.7^\circ \text{ and } \angle{C} = \omega \approx 18.4^\circ$ 
March 23rd, 2019, 08:45 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133  If about 5.6 were the smallest value, the smallest integer value would be 6. The smallest integer value for $BD$ is 5, as demonstrated in the diagram below. Figure1.PNG 
March 23rd, 2019, 09:37 AM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,580 Thanks: 1038 
Ahhh....gotcha! And angle ADB = 4w, of course...


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