Second Method: Trigonometry | Calculate the length AB? |

[@PreMath]

First method (Without Trigonometry) link: kzbin.info/www/bejne/kIHOZZx3h9ufgtU

[@alster724]

It's easier with trigonometry. Thanks!

[@marcelowanderleycorreia8876]

Very cool professor!!

[@PreMath]

Thanks for the feedback ❤️

[@phungpham1725]

Impressive method! Thank you so much!

[@phungcanhngo]

Thank you so much.

[@jamestalbott4499]

Thank you!

[@jimlocke9320]

Solution by similar triangles: Drop a perpendicular from B to AC and label the intersection as point E. Extend AC up to the right and drop a perpendicular to it from D, labelling the intersection as point F. Consider right ΔADF. It is a special 30°-60°-90° right triangle. Its hypotenuse AD has length 28, so AF, opposite the 30° angle, has length 14 and DF has length 14√3. CF = AC + AF = 70 + 14 = 84. Consider right ΔABE. It is also a special 30°-60°-90° right triangle. Let AB have length x, then AE = x/2, BE = x(√3)/2 and CE = AC - AE = 70 - x/2. Now note that ΔCDF and ΔCBE are similar by angle - angle (common angle

[@swadhinnayak469]

Well done 👍

[@LuisdeBritoCamacho]

This is the same Problem Given 3 days ago!!

[@UmaMohan-x4x]

Prof repeats not only the problems but the comments as well. His favourite is Let us go ahead.

[@marcgriselhubert3915]

Already seen and discussed.

[@nilsalmgren4492]

I think I like using law of cosines, then law of sines to find missing exterior angle. Then find the angles at b and use law of sines in one of the two smaller triangles to find x. I like proportions and it is very direct.

[@Irishfan]

I use Trigonometry to solve all my triangle problems. I also use the decimal value for the trig function and sides rather than the radical.

[@harikatragadda]

Extend AB to AE such that ∆AED is equilateral. ∆BAC is Similar to ∆BED. Hence, AB/70 = (28-AB)/28 AB = 20

[@PrithwirajSen-nj6qq]

With the help of Eisenstine theorem CD^2= 70*70+70*28+28*28 Getting CD we may get BC and BD with the help of angle bisector theorem. BC/BD=70/28=5/2 Then again by Eisenstein theorem 70*70-70,AB+AB^2=BC^2 BC known. Solving this equation we may get AB

[@unknownidentity2846]

I like that solution very much. Unfortunately it works that fast only for angles of 120°=2*60° because sin(120°)=sin(60°). Maybe you would like to create a third video using the angle bisector theorem to solve this problem. Best regards from Germany

[@quigonkenny]

It would work for other values of the angles, but it would only be so simple as long as the sine values were well known/simple to calculate (30°, 45°, and multiples thereof). If the greater angle was 135°, for instance, and the inner angles were 60° and 75°, the same procedure could be used, but if you didn't know the exact value of sin(75°) (it's (√2+√6)/4), you would either need to determine it from sin(45°+30°) or use a calculator. 120° and 60° just makes it particularly easy because both sine values are the same and well known, so the radicals cancel out. As for using the angle bisector theorem, I used that to solve it (along with the law of cosines and Stewart's Theorem) on the previous video for this problem.

[@unknownidentity2846]

@@quigonkenny Concerning the other angles you are absolutely right. Nonetheless the choice of 60°/60° is so attractive since it leads to the very simple and therefore beautiful result AB=AC*AD/(AC+AD). I also used the angle bisector theorem in combination with the law of cosines when I saw the first video. Since the angle bisector theorem was used not long time ago for another problem on this channel I was wondering why it was not used here. Best regards from Germany

[@Asphalt888.8]

This is the easiest method and I used this method to solve the problem.

[@quigonkenny]

I solved this on the previous video via lww of cosines (to determine CD), angle bisector theorem (to determine CB and DB), and Stewart's theorem (to determine AB).

[@PrithwirajSen-nj6qq]

By Eisenstein theorem from 🔺 ACD we say CD^2=AC^2+AC.AD+AD^2 Can derive CD. Then using angle bisector theorem we say CB/BD=70/28 (CB+BD)/BD =(70+28)/28 CD/BD=98/28=7/2 BD =CD*2/9 Now BD known In 🔺 ABD AD^2-AD*AB+AB^2=BD^2 by Eisenstein theorem. Here from we may get AD. (no construction required in the method)

[@sahyadrimedicals5683]

we can directly apply formula for length of angle bisector that is 2ab/a+b*cos(theta/2)

[@johnbrennan3372]

CB/ BD= 70/28= 10/4. Let |CB| = 10a and |BD|= 4a. Cos 120 degrees= 70^2+28^2 -(14a)^2 divided by 2(70)(28). That gives a= sqroot 39. |AB|^2= |AC||AD|- |CB||BD|(theorem). So |AB|^2= 1960- 1560. Therefore |AB|=20

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال DC^2=AC^2+AD^2-2AD.AC cosCAD BC/BD =AC/AD CD=14(racine39) CB=10(racine39) BD=4(racine39) BC^2=AC^2+x^2-2xACcos60 BD^2=AD^2+x^2-2xADcos60 ... x=20

[@kevingeier3385]

How does this help determine the length of x. Which was the question. This is not what the question asked.

[@comdo777]

asnwer=21cm isit

[@devondevon4366]

20 Law of cosine using 70, 120 degrees, and 28 This will yield 16.102 degrees for the smallest angle Then, using the Law of Sine using 16.102, 70 length, and 60 degrees. Hence, the unknown length = 20. Answer