Can you find the value of x? | Law of Cosines |

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Learn how to find the value of x. Important Geometry and Algebra skills are also explained: Inscribed angles theorem; Law of cosines; Isosceles triangles. Step-by-step tutorial by PreMath.com.
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9 окт 2024

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Комментарии : 45

@RAG981 6 месяцев назад

Lovely simple result. How unusual for the centre of the circle not to be relevant.

@PreMath 6 месяцев назад

Yes, it is! Thanks for sharing ❤️

@mhcheung8392 5 месяцев назад

BCD is an equilateral triangle. From triangle ACD, by cosine law, CD=sqrt91. Therefore, BC=BD=CD=91. Consider triangle ABC, by cosine law again, a quadratic equation x^2-6x-55=0 would be set up. So, x=11

@wackojacko3962 6 месяцев назад

Absolutely love it! 🙂

@PreMath 6 месяцев назад

Glad you like it! Thanks ❤️

@Abby-hi4sf 6 месяцев назад

Great tutorial!

@marcgriselhubert3915 6 месяцев назад

As often I use an orthonormal. I choose center A and first axis (AB) As angleBAC = 60° we have C(6.cos(60°); 6.sin(60°)), so C(3; 3.sqrt(3)). In the same manner we have D(5/2; (-5/2).sqrt(3)), and naturally A(0;0) The equation of the circle is x^2 + y^2 +a.x +b.y +c = 0. As A is on the circle, then c = 0, so the equation is x^2 + y^2 +a.x + b.y = 0 B is on the circle, then 9 +27 +3.a +3.sqrt(3).b = 0, or a + sqrt(3).b = -12. C is on the circle, then 25/4 + 75/4 +(5/2).a -(5/2).sqrt(3).b = 0, or a -sqrt(3).b = -10 Then we get a = -11 and b = -1/sqrt(3) = -sqrt(3)/3. The equation of he circle is x^2 + y^2 -11.x -(sqrt3)/3).y =0 The intersection of the circle and the axis (AB), which equation is y = 0, gives that the abscissa of B is 11, and finally the distance x = AB = 11.

@PreMath 6 месяцев назад

Thanks for sharing ❤️

@soli9mana-soli4953 6 месяцев назад

first I found DC with cosine law then found CH and DH with bisector theorem (H is the intersection of line CD with line AB) then found AH² = AC*AD - CH*DH (stewart theorem?) then fond BH with intersecting chord theorem BH * AH = CH*HD and finally AB = AH+BH = 91/11 +30/11=121/11=11

@ybodoN 6 месяцев назад

AH² = AC · AD − CH · DH also follows from the angle bisector theorem: “the length of the angle bisector _d_ can be found by _d² = bc − mn”._ But this can be proven by Stewart's theorem and there is nothing wrong to get there by a longer way which should earn bonus points 😉

@sandanadurair5862 6 месяцев назад

In a cyclic quadrilateral Product of diagonals = sum of products of opposite sides AB.DC = AC.BD+AD.BC m.x = 6.m + 5.m m.x = 11.m X = 11 units AB=x BD=BC=DC=m AC = 6 AD = 5

@PreMath 6 месяцев назад

Thanks for sharing ❤️

@Okkk517 6 месяцев назад

It is a nice solution. However, it is applicable for this case (60 deg.) only. I wonder how the theorem could be used in the general case.

@sandanadurair5862 6 месяцев назад

@Okkk517 I agree with you. The relationship holds good for any cyclic quadrilateral. However, in this particular case , we could conclude BC, BD, and CD sides are of equal length. Hence we could find the solution easily. Thank you for your feedback

@ybodoN 6 месяцев назад

Connect CB and DB. Let O be the center of the circle. By the inscribed angle theorem, COB = DOB = 120° therefore CB = DB. By the law of cosines, we can write the equation system {y² = 6² + x² − (2) (6) (x) cos(60°), y² = 5² + x² − (2) (5) (x) cos(60°)}. Since cos (60°) = ½ then we can rewrite it as {y² = 6² + x² − 6x, y² = 5² + x² − 5x}. The only positive solution is y = √91, x = 11. Now let's watch the video! 🤓 Thank you PreMath! 🙏

@PreMath 6 месяцев назад

You are very welcome! Thanks for sharing ❤️

@prossvay8744 6 месяцев назад

Connect B to C and B to D By angle BAC=BAD=60° So BC=BD=a Cos(60°)=(36+x^ 2 -a^2)/12x 1/2=(36+x^2-a^2)/12x a^2=x^2-6x+36 (1) Cos(60°)=(25+x^2-a^2)/10x 5x=x^2+25-a^2 a^2=x^2-5x+25 (2) (1) and (2) x^2-6x+36=x^2-5x+25 So x=11units.❤❤❤ Thanks sir.

@PreMath 6 месяцев назад

Excellent! You are very welcome! Thanks for sharing ❤️

@unknownidentity2846 6 месяцев назад

Let's find x: . .. ... .... ..... Let's assume that the center of the circle with the radius R is also the center of the coordinate system and that AB is parallel to the x-axis. In this case we obviously have xA=−xB and yA=yB and therefore x=AB=2xB. Since A, C and D are located on the circle, we can conclude: xA² + yA² = R² xC² + yC² = R² xD² + yD² = R² xA² + yA² = R² [xA + AC*cos(60°)]² + [yA + AC*sin(60°)]² = R² [xA + AD*cos(60°)]² + [yA − AD*sin(60°)]² = R² xA² + yA² = R² [xA + 6*(1/2)]² + [yA + 6*(√3/2)]² = R² [xA + 5*(1/2)]² + [yA − 5*(√3/2)]² = R² xA² + yA² = R² (xA + 3)² + (yA + 3√3)² = R² (xA + 5/2)² + (yA − 5√3/2)² = R² xA² + yA² = R² xA² + 6xA + 9 + yA² + 6√3yA + 27 = R² xA² + 5xA + 25/4 + yA² − 5√3yA + 75/4 = R² 6xA + 6√3yA + 36 = 0 5xA − 5√3yA + 25 = 0 xA + √3yA + 6 = 0 xA − √3yA + 5 = 0 2xA + 11 = 0 ⇒ xA = −11/2 ⇒ xB = 11/2 ⇒ x = AB = 2xB = 11 Best regards from Germany

@PreMath 6 месяцев назад

Excellent! Thanks for sharing ❤️

@santiagoarosam430 6 месяцев назад

Llamamos F y D a las proyecciones ortogonales de C y de D sobre AB. El segmento que une C y D corta a EF en el punto M. Si AC=6 y AD=5 → AF=3 ; CF=3√3 ; AE=5/2 ; ED=5√3/2 → EF=1/2 =EM+MF =EM+[(1/2)-EM] → EM/ED=MF/CF→ EM=5/22→ MF=3/11 → CM²=MF²+CF²→ CM=6√91/11 y MD²=ME²+ED²→ MD=5√91/11. Potencia de M respecto a la circunferencia propuesta = (AE+EM)(MF+AB-AF)=CM*MD → [(5/2)+(5/22)][(3/11)+X-3]=(6√91/11)(5√91/11)→ X=11 Interesante problema. Gracias y saludos.

@DB-lg5sq 6 месяцев назад

شكرا لكم على المجهودات يمكن استعمال SinADB / x = sinCAD / DC = 1/2R CD= racine 91 ...... x=11

@chileunglau4423 5 месяцев назад

I try the other approach to solve this question. First, draw a chord DE passing through D and parallel to AC which cuts AB at F. Connecting BE and CE you will have two equilateral triangles and a parallelogram. Then you can find that AB = AC + AD

@jimlocke9320 6 месяцев назад

It can be solved using the intersecting chords theorem. Drop a perpendicular from C to AB and label the intersection E. Drop a perpendicular from D to AB and label the intersection F Construct CD and label the intersection with AB as G. ΔACE is a 30°-60°-90° special right triangle, AE = 3 and CE = 3√3. ΔADF is a 30°-60°-90° special right triangle, AF = 2.5 and DF = 2.5√3. ΔCEG and ΔDFG are similar, corresponding sides CE and DF are in the ratio 3√3:2.5√3 = 3:2.5 = 6:5. So sides of ΔCEG are 6/5 times as long as corresponding sides of ΔDFG. So, since FE = AE - AF = .05, FG (5/11)(0.5) and EG = (6/11)(0.5). We can work out length CD = √(91) (construct a larger right triangle ΔCDH with sides DH = 0.5 and CH = 5.5√3 and compute the hypotenuse) and divide it proportionally into CG and DG. Then we add AF and FG to find AG and apply intersecting chords theorem to AB and CD ( (AG)(BG) = (CG)(DG) to find BG. X = AB = AG + BG. It can be done without a calculator but PreMath's method is much more straightforward!

@podgem-l4t 6 месяцев назад

Yep, I had the same idea, the second a problem says sin/cos ... my mind immideatley goes to, How do I solve this without using that

@PreMath 6 месяцев назад

Thanks for sharing ❤️

@jamestalbott4499 6 месяцев назад

Thank you!

@thinker821 5 месяцев назад

This can be solved with trigonometry. For triangle ACD, you have two sides (5 and 6) and the angle contained by them (120°). Find the circumradius R as √91/√3 using this information. Let O be the centre of the circle. Then find the cosine of the angle CAO using the chord length AC = 6 and the circumradius. It will be 3√3/√91. The sine of the angle is therefore 8/√91. Then angle OAB = 60° - angle CAO, then find the chord length as 2*R*cos(60°- OAB). The answer will be 11.

@dumitruolteanu8663 6 месяцев назад

Or,with the Van Shooten's theorem( particullarity of Ptolemeu's theorem): x=6+5

@georgexomeritakis2793 5 месяцев назад

You can solve this by drawing 2 equilateral triangles with common side AB and compare the arcs that are formed on the circle

@MrPaulc222 6 месяцев назад

I took a totally different direction but wasn't sure if I had the right answer (I did), but I am unclear if this would work in all instances of a similar puzzle. I would appreciate some feedback on this matter please. I reasoned that if x were the diameter, the sides providing the 30,60,90 triangles would average out at 5.5, making the hypotenuse (x), 11, but maths is rarely that clean. With x as the diameter, x would be 12 if you use 6 as the short side of a 30,60,90 and x would be 10 if you use 5 as the short side of a 30,60,90. I didn't actually expect the average (11) to be correct, but I did expect it to be close. Have I stumbled across something here, or was I just lucky because the numbers worked out?

@misterenter-iz7rz 6 месяцев назад

Very difficult, but I can compute the radius of the circle, as CD^2=36+25-60cos 120=36+25+30=91, then (sqrt(91)/2)/r=sin 60=sqrt(3)/2， r=2sqrt(91)/2 sqrt(3)=sqrt(91/3) 😊

@misterenter-iz7rz 6 месяцев назад

Continue the solution😅, cos a=3/(sqrt(91/3)), sin a=sqrt(1-9/(91/3))=sqrt(1-27/91)=sqrt(64/91)=8/sqrt(91), cos b=cos (60-a)=cos 60 cos a+sin 60 sin a=1/2 3/(sqrt(91/3))+sqrt(3)/2 8/sqrt(91)=3sqrt(3)/2sqrt(91)+4sqrt(3)/sqrt(91)=11sqrt(3)/2sqrt(91), x=2r. cos b=2 sqrt(91/3) 11sqrt(3)/2sqrt(91)=2 sqrt(91)/sqrt(3) 11sqrt(3) / 2 sqrt(91)=22sqrt(91)sqrt(3)/2sqrt(91)sqrt(3)=11.😅😅😅😅😅😅😅😅😅

@PreMath 6 месяцев назад

Thanks for sharing ❤️

@misterenter-iz7rz 6 месяцев назад

@PreMath My solution is awful😢, yours likes a magic show.🎉

@PreMath 6 месяцев назад

@@misterenter-iz7rz No worries. We are all lifelong learners. 😀🌹

@riteshkumar8152 6 месяцев назад

Your teaching method is Nice sir. Plss pin

@PreMath 6 месяцев назад

Keep watching🌹 Thanks ❤️

@BKNeifert 6 месяцев назад

Can you do a puzzle with a 30 60 90 right triangle, but solve using radians? Like you did with that Japanese one, it helped me understand sin and cosine. Thanks. Also, keep me anonymous, please.

@eugeniolopez7572 4 месяца назад

< DCB=60° and < CDB= 60° CBD = 60° then CD = CB = BD = m

@s900991 5 месяцев назад

@AmjadAli-lk3pf 5 месяцев назад

Sir,any contact I have a one question.

@Toztabud 6 месяцев назад

Is that shape even valid? BCD is an equilateral triangle. Angle CAD is 120 degrees and AB is supposed to evenly bisect the angle? Yet AC is 6 units and AD is 5 units? That's not a valid shape, so there is no answer.

@ybodoN 6 месяцев назад

However, as many others on this channel, I have absolutely no problem drawing it perfectly to scale… 🤔 Points A, C and D are sufficient to define the circle. Then bisect ∠CAD to find B on the circumference 😉

@subs111222 5 месяцев назад

Bit simpler .. Is to draw a diameter AE from point A.. Joining CE and DE and EB forms 3 right angled triangles.. Considering angle # between AE and x, angle CAE becomes 60-# and angle DAE becomes 60+#.. Applying cosine on triangles CAE and DAE you get cos#=11/AE.. From 3rd triangle AEB, you get cos#=x/AE.. Thus x equals 11