Can you find the side length X? | Isosceles triangle | (Important skills explained) |

[@bigm383]

Thanks, Professor, for an excellent problem!❤

[@PreMath]

You are very welcome! Thanks for your feedback! Cheers! 😀 You are the best❤️. Keep it up 👍

[@bigm383]

@@PreMath😂😀

[@HappyFamilyOnline]

Amazing 👍 Thanks for sharing 😊

[@CharlesB147]

That is a clever little solution trick, using a double sine to set the equations exactly equal to each other.

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!!!!!

[@harikatragadda]

Draw a circle with C as the center passing through A and B. Extend AC to E and AD to G on the circle to make a right triangle AGE. Drop a perpendicular CF on AG, and since ∠CDF = 30° and CD= 4, we have CF = 2. By proportion, EG = 2CF = 4. We now have ∆ABE Congruent to ∆AGE since ∠AEB = ∠EAG = 24° and their Hypotenuse is equal to the diameter. Hence EG = AB = X = 4

[@timeonly1401]

Wow!! Nicely done! I LOVE geometry-only solutions. For almost all these problems, I haven't been able to see them (YET!), but I appreciate them when I see others present them! I'm working on it! LOL.. 😅

[@wackojacko3962]

@ 5:27 the perpendicular CE on AB is the perpendicular bisector of the base of isosceles triangle CAB and also the angle bisector of the vertex angle ACB. A beautiful isosceles angle theorem! 🙂

[@PreMath]

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

[@arnavkange1487]

You are amazing

[@PreMath]

Thank you! Cheers! 😀 You are awesome. Keep it up 👍

[@jimlocke9320]

Solution by construction (no trigonometry). Extend AD to the right. Drop a perpendicular to the extended AD and call the intersection F. Note that, at 3:27, PreMath determines that

[@sumithpeiris8440]

Easy Pure Geometry Solution Let H be the midpoint of AB, let AD meet AH at E, BC at F and the perpendicular to AF from C at G Triangle CDG is a 30-60-90 Triangle, so CG = 4/2 = 2 Now

[@zupitoxyt]

It need a diagram

[@arnavkange1487]

This law of sines was new for me

[@PreMath]

Excellent! You are awesome. Keep it up 👍

[@mibsaamahmed]

It's nice that you learned something new

[@th3smurf692]

It's an important law 👍

[@Copernicusfreud]

Yay!, I solved the problem.

[@AndreyDanilkin]

ACB - 66;48;66. ADC - 24;150;6. Let's flip the triangle ADС along side AC (change points A and С). Now

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@dannymeslier6658]

Elegant solution, cheers

[@murdock5537]

Excellent, many thanks, Sir! φ = 30°; α = 42°; β = 24° → α + β = θ = 66° → BCA = 2β → DCA = β/4 → 6φ - (β + β/4) = 5φ → sin⁡(5φ) = sin⁡(6φ - 5φ) = sin⁡(φ) = 1/2 AC = BC = a → CAB = ABC = θ → BCA = 6φ - 2θ = 2(3φ - θ) = 2β AC = BC = a; AB = x → x/2 = AE = BE → sin⁡(φ)/a = 1/2a = sin⁡(β)/4 → sin⁡(β) = 2/a sin⁡(β) = (x/2)/a = x/2a = 2/a → x = 4 btw: cos⁡(θ) = (x/2)/a = x/2a = 2/a → a = 2/cos⁡(θ) ≈ 4,92 AD ∶= k = ? sin⁡(β/4)/k = sin⁡(β)/4 = sin⁡(φ)/a = 1/2a → sin⁡(β/4) = k/2a → k = 2asin⁡(β/4) = 4sin⁡(β/4)/cos⁡(θ) = 4sin⁡(β/4)/sin⁡(β) ≈ 1,03 or: CE = EM + CM ↔AM = AD + DM → AM = CM = r φ = 30°; α = 42°; β = 24° → α + β = θ = 66° → CAB = ABC = θ → BCA = 6φ - 2θ = 2(3φ - θ) = 2β; BCD = α → ECD = α - β = 3β/4 → DCA = β/4 → ADC = 6φ - 5β/4 = 5φ → sin⁡(5φ) = sin⁡(6φ - 5φ) = sin⁡(φ) = 1/2 CAM = MCA = β → AMC = 6φ - 2β = 132° → sin⁡(6φ - 2β) = sin⁡(6φ - (6φ - 2β)) = sin⁡(2β) → sin⁡(2β)/4 = sin⁡(φ)/r = 1/2r → r = 2/sin⁡(2β) ≈ 2,69 cos⁡(α) = (x/2)/r = x/2r → x = 2rcos(α) = 4 or (no calculator needed): φ = 30°; α = 42°; β = 24° → α + β = θ = 66° → 6φ - 2θ = 2β AC = BC = a → CAB = ABC = θ → BCA = 2β → BCE = ECA = β BCD = α → ECD = β - α = 3β/4 → DCA = β/4 → ADC = 6φ - 5β/4 = 5φ → sin⁡(5φ) = sin⁡(6φ - 5φ) = sin⁡(φ) = 1/2 CE = CM + EM → AM = AD + DM = CM = r ∆DMC → CDM = φ → MCD = 3β/4 → DMC = 6φ - (φ +3β/4) = 2θ → sin⁡(2θ)/4 = sin⁡(φ)/r = 1/2r → r = 2/sin⁡(2θ) → sin⁡(2θ) = sin⁡(6φ - 2θ) = sin⁡(2β) → r = 2/sin⁡(2β) α + 2β = 3φ → sin⁡(α) = cos⁡(2β) → cos⁡(α) = sin⁡(2β) AE = BE = x/2 → cos⁡(α) = (x/2)/r = x/2r → x = 2rcos(α) = 4cos⁡(α)/sin(2β) = 4cos⁡(α)/cos⁡(α) = 4

[@johnbrennan3372]

Find length “a” by sine rule and then use cosine rule in triangle ABC.

[@mibsaamahmed]

Can someone tell me an interesting tooic related to geometry or trigonometry to study

[@marioalb9726]

Isósceles triangle: β = 44° + 22° = 66° α = 180° - 2 . 66° = 48° Internal triangle: γ = 48° - 42° = 6° θ = 180° - 6° - 24° = 150° Sine rule: a / sin 150° = 4 / sin 24° a = 4,917 cm x = 2 . a. cos 66° x = 4 cm ( Solved √ )

[@devondevon4366]

x=4 Angle B = 66 (since the triangle is an isosceles) Angle C = 48 (180 - 132) Hence the angles of triangle ADC: 6, 150, and 24 (degrees). Since you now have Angle Side Angle: 6, 4, and 150, you can use the sine formulae a/sine a = b/sine B to get the length of AC = 4.917. Hence the length of BC= 4.917 ( since the triangle is an isosceles) Since you again have Angle Side Angle: 48 degrees, 4.917 unit, and 66 degrees, you can again use the sine formulae to get x=4 Answer

[@MarieAnne.]

At 0:11 you mention that ∠ADC = 24°. I think you meant ∠DAC.

[@じーちゃんねる-v4n]

From the law of sine AC=4sin150/sin24 ∴x=2ACcos66=8sin30cos66/sin24=4

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@Golololololo]

Thanks for your helpful videos Can I send you a question, sir? If yes, how?

[@giuseppemalaguti435]

x=4sin30sin48/sin24cos24=4*1/2*2=4

[@devondevon4366]

4

[@unknownidentity2846]

Ok, lets have a try: . .. ... .... ..... AC = BC ⇒ ∠BAC = ∠ABC = 24° + 42° = 66° ⇒ ∠ACB = 180° − ∠ABC − ∠BAC = 180° − 66° − 66° = 48° ⇒ ∠ACD = ∠ACB − ∠BCD = 48° − 42° = 6° ⇒ ∠ADC = 180° − ∠CAD − ∠ACD = 180° − 24° − 6° = 150° AC/sin(∠ADC) = CD/sin(∠CAD) AC/sin(150°) = 4/sin(24°) AC/(1/2) = 4/sin(24°) AC = 2/sin(24°) AB/sin(∠ACB) = AC/sin(∠ABC) x/sin(48°) = (2/sin(24°))/sin(66°) x = 2*sin(48°) / [sin(24°)*sin(66°)] x = 2*sin(2*24°) / [sin(24°)*cos(90°−66°)] x = 2*2*sin(24°)*cos(24°) / [sin(24°)*cos(24°)] x = 4 Best regards from Germany

[@JSSTyger]

x = 4

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

[@축복이-x6u]

asnwer=12cm you tube ad isit