 My Math Forum How to find the smallest length in a triangle when the angles aren't known?
 User Name Remember Me? Password

 Geometry Geometry Math Forum

 March 21st, 2019, 05:44 PM #1 Senior Member   Joined: Jun 2017 From: Lima, Peru Posts: 114 Thanks: 2 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: 3,044 Thanks: 1627 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: 14,597 Thanks: 1039 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
Senior Member

Joined: Jun 2017
From: Lima, Peru

Posts: 114
Thanks: 2

Math Focus: Calculus More help please

Quote:
 Originally Posted by skeeter 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"
I've also tried to follow this method but the part where I got stuck was at:

$\frac{4}{\cos 2\omega}= BD$

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
Senior Member

Joined: Jun 2017
From: Lima, Peru

Posts: 114
Thanks: 2

Math Focus: Calculus
Quote:
 Originally Posted by Denis 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.
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: 3,044
Thanks: 1627

Quote:
 Originally Posted by Chemist116 I've also tried to follow this method but the part where I got stuck was at: $\frac{4}{\cos 2\omega}= BD$ 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? $\dfrac{4}{\cos(2\omega)} = |BD| \implies \dfrac{4}{|BD|} = \cos(2\omega)$

$0 < \cos(2\omega) < 1$

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: 14,597
Thanks: 1039

Quote:
 Originally Posted by Chemist116 Find the smallest whole number that BD can attain.
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: 3,044 Thanks: 1627 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: 21,036
Thanks: 2273

Quote:
 Originally Posted by Denis ~5.656.... thus 5 as per Skeeter.
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: 14,597 Thanks: 1039 Ahhh....gotcha! And angle ADB = 4w, of course... Tags angles, find, length, smallest, triangle Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Larrousse Trigonometry 4 April 22nd, 2018 05:11 PM adilgrt007 Algebra 9 August 22nd, 2013 10:25 AM mathmaniac Algebra 16 February 14th, 2013 06:27 PM MHSK Algebra 2 February 18th, 2012 04:00 AM billybob Calculus 5 May 12th, 2011 02:37 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      