Nice sir, I'm your big fan

@PreMath

2 ай бұрын

Thanks dear❤️

Wow, amaze!

You made it unbelievable simple

Let the height = h. h/8 = tan x h = 8.tan x h/2 = tan 2x h = 2.tan 2x We get 8.tanx = 2.tan 2x 4.tan x = tan 2x Knowing tan 2x = 2.tan x/(1-tan²x) So 4.tan x = 2.tan x/(1-tan²x) 1-tan²x = 2.tan x/4.tan x 1-tan²x = ½ 1-½ = tan²x ½ = tan²x tan x = ±√(1/2) = ±½√2 Since x is acute, then tan x = ½√2 h/8 = ½√2 h = 8.½√2 = 4√2 Using phytagorean for determining both hypotenuses, and we've done.

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

@robith100

2 ай бұрын

@@PreMath love ur channel 😍

Hello! I fought hard, because I didn't see the much easier construction of auxiliary lines, I constructed an angle bisector from point B. Too complicated to describe, to type everything in here, but I found relations leading to three equations with 3 unknowns, finally and luckily found the values we were looking for - correctly. But your shown way was much easier and much faster. Why didn't I see that myself? I hope I will get better, but at least I am glad, that I did it myself and I didn't have to give up. BTW, I am speechless about how many videos you produce and put them on line. When I solved one, the next one is already waiting. My deep respect and thank you very much!

use of formula - tan2A= 2tanA/1-tan^2A- makes it much easier.

@PreMath

2 ай бұрын

Thanks for sharing ❤️

AMAZING! Seeing some of your solutions is like seeing magic. Thank you so much for providing this wonderment!

By definition of acute angles in a right triangle, m∠ACD = (90 - x)°. Draw a segment thru point C and a point E on segment AD, such that △BCE is an isosceles triangle. So, BC = CE, BD = DE = 2, & m∠BEC = (2x)° by the Base Angles Theorem. Then, AE = 8 - 2 = 6 by the Segment Addition Postulate. Again, by definition of acute angles in a right triangle, m∠DCE = (90 - 2x)°. Use the Angle Addition Postulate. m∠ACD = m∠ACE + m∠DCE (90 - x)° = m∠ACE + (90 - 2x)° m∠ACE = (90 - x)° - (90 - 2x)° = 90 - x - 90 + 2x m∠ACE = x° Therefore, △AEC is an isosceles triangle and AE = CE by the Base Angles Converse. By the Transitive Property of Equality (or Segment Congruence), BC = 6. One down, two to go! Apply the Pythagorean Theorem on △BDC (or △EDC). a² + b² = c² (CD)² + 2² = 6² (CD)² + 4 = 36 (CD)² = 32 CD = √32 = (√16)(√2) = 4√2 For the last side, use the Pythagorean Theorem on △ADC. c² = a² + b² (AC)² = (4√2)² + 8² = 32 + 64 = 96 AC = √96 = (√16)(√6) = 4√6 Thus, the lengths of the missing sides are as follows: AC = 4√6 units ≈ 9.80 units BC = 6 units CD = 4√2 units ≈ 5.66 units

Let's do it: . .. ... .... ..... Since the triangles ACD and BCD are right triangles, we can conclude: tan(x) = CD/AD ∧ tan(2x) = CD/BD ⇒ tan(2x)/tan(x) = (CD/BD)/(CD/AD) = (CD/BD)*(AD/CD) = AD/BD = 8/2 = 4 tan(2x) = 4*tan(x) 2*tan(x)/[1 − tan²(x)] = 4*tan(x) 1/2 = 1 − tan²(x) tan²(x) = 1/2 ⇒ tan(x) = 1/√2 = √2/2 CD/AD = tan(x) ⇒ CD = tan(x)*AD = (√2/2)*8 = 4√2 Now we can apply the Pythagorean theorem to the right triangles ACD and BCD: AC² = AD² + CD² = 8² + (4√2)² = 64 + 32 = 96 ⇒ AC = √96 = 4√6 BC² = BD² + CD² = 2² + (4√2)² = 4 + 32 = 36 ⇒ BC = √36 = 6 Best regards from Germany

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

Refreshing! 🙂

@PreMath

2 ай бұрын

Glad to hear that! Thanks ❤️

Tan(x)=h/8 CD=h Tan(2x)=h/2 Tan(2x)=2tan(x)/1-tan(x)^2=h/2 2 h/8/1-(h/8)^2=h/2 So h=4√2 AC=√64+(4√2)^2=4√6 units CD=h=4√2 units BC=√2^2+(4√2)^2=6 units.❤❤❤ Thanks sir.

@PreMath

2 ай бұрын

Excellent! You are very welcome! Thanks for sharing ❤️

Nice!

Superb!👍

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

Draw CE, where E is the point on AD where ED = 2. As ED = DB, ∠EDC = ∠CDB, and CD is shared, ∆EDC and ∆CDB are congruent by SAS and thus ∠CED = ∠DBC = 2x. As ∠CED is an external angle to ∆AEC at E, ∠ECA = ∠CED - ∠CAE = 2x-x = x. As ∠ECA = ∠CAE = x, ∆AEC is isosceles and EC = AE = 8-2 = 6. As ∠EBC = ∠CEB = 2x, ∆BCE is isosceles and BC = CE = 6. BC = 6 ✓ Triangle ∆CDB: DB² + CD² = BC² 2² + CD² = 6² CD² = 36 - 4 = 32 CD = √32 = 4√2 CD = 4√2 ✓ Triangle ∆ADC: DC² + AD² = CA² (4√2)² + 8² = CA² CA² = 32 + 64 = 96 CA = √96 = 4√6 CA = 4√6 ✓

Nice Solution. But small correction the 2 triangles are congruent by AAS not by ASA.

Using tan x and tan 2x and equating the two values for cd works out easy giving tan x = 1/sqroot2. So cd/ 8= 1/sqroot2 etc

@PreMath

2 ай бұрын

Thanks for sharing ❤️

this one was easy thanks to your amazing explanation , I really appreciate . I suggest also calculus of Maclaurin and Taylor series if possible . thanks to all your video lessons I understand many stuffs in geometry . Thanks Sir

I really appreciate your video lessons, it really make my days happier, Could you mind Telling us your name. I express my sincere gratitude to you, Sir.

Very nice solution. Thank you.

@PreMath

2 ай бұрын

You are very welcome! Thanks ❤️

I like to see answers in decimal form. CB = 5.6569 AC = 9.7980 note I show equals sign because my tablet doesn't have an approximate sign. I carry them out to the nearest ten thousandth just for further use in calculations. The work l do in construction and civil engineering usually doesn't require final answers in a degree of accuracy that high. When I do calculations using a calculator, I leave the preliminary answers in memory carried out as far as the calculator will display. When I look at a number that is 5.66 units I know how long that is, when I look at 4 x the square root of 2, I have no idea how long that is and I have to get out my calculator convert it to decimal form. So, to me, when a math instructor wants an answer simplified, the simplest form wouldn't be an answer with a square root in it. The simplest form would be a digital number.

I got two answers, [yellow, middle, blue] =[4sqrt10, 4sqrt6, 10] or [4sqrt6, 4sqrt2, 6]. They check when a2+b2=c2 for the yellow and blue triangles, but the angles dont check for the first set, so the second set is correct.

شكرا لكم على المجهودات يمكن استعمال tanx=h/8 tan2x=h/2 ...... h=4 (racine 2)

1) tan(X) = CD/8 2) tan(2X) = CD/2 3) CD = 8*tan(X) 4) CD = 2*tan(2X) 5) 8*tan(X) = 2*tan(2X) 6) tan(2X) = 2*tan(X) / (1 − tan^2(X)) 7) CD = h 8) h = 8*tan(X) 9) h = 2*tan(2X) 10) 8*tan(X) = 2*tan(2X) 11) 8*tan(X) = 4*(tan(X) / (1 - tan^2(X)) 12) 8*(1 - tan^2(X) = 4 13) 1 - tan^2(X) = 1/2 14) 1 - (h/8)^2 = 1/2 15) 1 - h^2/64 = 1/2 16) 64 - h^2 / 64 = 1/2 17) 64 - h^2 = 32 18) h^2 = 64 - 32 19) h^2 = 32 20) h = sqrt(32) = 4*sqrt(2) ~ 5,7 lin un 21) X = arctan(4*sqrt(2)/8) ; X = arctan(sqrt(2)/2) ; X ~ 35,3º 22) 2X = arctan(4*sqrt(2)/2) ; 2X = arctan(2*sqrt(2)) ; 70,5º 23) AC^2 = 64 + 32 ; AC^2 = 96 ; AC = sqrt(96) ; AC = 4*sqrt(6) ; AC ~ 9,8 lin un 24) BC^2 = 4 + 32 ;BC^2 = 36 ; BC = 6 lin un 25) My Best Answer is that AC is equal to 4*sqrt(6) Linear Units (approx. 9,8 Linear Units) and BC = 6 Linear Units.

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

In triangle ADC: tan(x) = DC/8, and in triangle BDC: tan(2.x) = DC/2, so we have tan(2.x) = 4.tan(x). Or tan(2.x) = (2.tan(x))/ (1 -tan(x)^2) So 4.tan(x) = (2.tan(x))/(1 -tan(x)^2) and 4.tan(x) - 4.tan(x)^3 = 2.tan(x), that gives tan(x)^2 = 1/2 and tan(x) = sqrt(2)/2 as tan(x) >0 Now we get sqrt(2)/2 = DC/8 and DC = 4.sqrt(2). Then in triangle ADC: AC^2 = 8^2 + (4.sqrt(2))^2 = 96 and AC = sqrt(96) = 4.sqrt(6) and in triangle BDC: BC^2 = 2^2 + (4.sqrt(2))^2 = 36 and BC = sqrt(36) = 6.

a/2=tan 2x, a/8=tan x, so 4=2/(1-tan^2 x), tan^2 x=1/2, a=8 tan x=8/sqrt(2), sec^2 x=3/2, cos^2 x=2/3, sin^2 x=1/3, b cos x=8, b=8/cos x=8sqrt(3)/sqrt(2), cos 2x=2×(2/3)-1=1/3, c cos 2x=2, c=2/cos 2x=2/(1/3)=6.

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

Much easier: bisector ВК ∠СВD. ▲АСD ~ ▲ВDК. DК/КС = 1/3. ВD/ВС = 1/3 (bisector property). ВС = 6. СD = 4√2, АС = 4√6 (Pythagorean theorem).

@PreMath

2 ай бұрын

Thanks for sharing ❤️

1:37 AAS congruency, not ASA

@rabotaakk-nw9nm

2 ай бұрын

1:10 construct: ED=BD=2, not SAS congruence =>

Pice of a cake. It is about inner and outer engels in a circle.

It can be easily solved isoscelizing....inside or outside 😂

2nd comment

@PreMath

2 ай бұрын

Thanks ❤️

Nice! φ = 30°; ∆ ABC → AB = AD + BD = 8 + 2 = 10; BC = a = ?; AC = b = ?; CD = h = ? sin⁡(ADC) = 1 → a = √(h^2 + 4); b = √(h^2 + 64); sin⁡(2δ) = h/a; sin⁡(δ) = h/b → cos⁡(δ) = 8/b → sin⁡(2δ) = 2sin⁡(δ)cos⁡(δ) → 16/b^2 = 1/a → 16a = b^2 → h^2 = a^2 - 4 = 16a - 64 → (a - 8)^2 = 4 → a1 = 10 = AB; a2 = 6 = a → h = 4√2 → b = 4√6 → sin⁡(δ) = √3/3 → δ ≈ 35,26°