My son's solution (He does not like Geometry) 😊 Δ ABD => tanθ=x/AB (1) Δ ABC => tanθ= ΑΒ/3x (2) (1),(2) => x/AB=AB/3x => AB=x√3 (2) => tanθ= x√3/3x = √3/3 => θ=30°
@quigonkenny
Ай бұрын
As ∠DAB = ∠BCA = θ and ∠B is common, ∆ABD and ∆ABC are similar triangles. AB/BC = BD/AB AB/3x = x/AB AB² = 3x² AB = √(3x²) = √3x tan(θ) = AB/BC = √3x/3x = 1/√3 θ = tan⁻¹(1/√3) = 30°
@ludmilaivanova1603
Ай бұрын
yes, I solved this the same method. Only the difference is in calculation of hypotenuse and then the sin of Theta.
@juanalfaro7522
24 күн бұрын
Let AB=Y, then X/Y = Y/3X -> Y^2 = 3*X^2 -> Y=X*sqrt (3) --> tan (DAB) = 1/sqrt (3) --> Theta = 30 degrees.
@haiduy7627
Ай бұрын
❤❤
@haiduy7627
Ай бұрын
🎉🎉❤❤🎉🎉
@michaeldoerr5810
Ай бұрын
Hey have you thought of a third method? I have noticed two or three comments talking abt a third method. I thought that the first method was fascinating in that it requires an exterior angle. Kind of.
@cosmolbfu67
Ай бұрын
x/AB = AB/3x AB^2 = 3•x^2 AB = sqrt(3)•x tanC = sqrt(3)•x / 3x = 1/sqrt(3) Thus, C = 30degree
@SViyaasJayavel
Ай бұрын
Short form Vertical height y then Tan@=x/y Tan@=y/3x Equating x/y is 1/√3 Therefore theta is 30°
@ManojkantSamal
15 күн бұрын
In triangle ABD tan*=BD/AB (*=read as thita) tan*=x/AB........ Eqn1 In triangle ABC tan*=AB/BC=AB/3X.... Eqn2 Comparing eqn1 &eqn2, we shall get the following X/AB=AB/3X (AB)^2=3.x^2 (^=read as to the power ) AB=#3.X(read as root over) In triangle ABC AC^2=AB^2 +BC^2 =(#3.X)^2+(3.X)^2 =3.X^2+9.X^2 =12.X^2 AC=#(12.x^2) =2.#3.x In triangle ABC Sin*=AB/AC =(#3.X)/(2.#3.X) =1/2 Sin*=Sin 30 *= 30 degree (may be) In triangle ABD tan*=BD/AB =X/(#3.X) =1/#3=tan 30 *=30 degree.....
@sergeyvinns931
Ай бұрын
30 degrees!
@kinno1837
Ай бұрын
3rd method: Angle BAD = Angle ACB = θ Angle ABD = Angle CBA = 90° (common angle) By AA similar, ∆ABD ~ ∆CBA Thus, AB/BD=BC/AB AB²=x•3x=3x² AB=√3x Thus, tanθ=x/√3x=1/√3 θ=30°
φ = 30° → sin(3φ) = 1; ∆ ABC → AB = a; BC = BD + CD = x + 2x = 3x; BCA = DAB = θ = ? tan(θ) = x/a = a/3x → a = x√3 → AD = 2x → sin(θ) = 1/2 = sin(φ) → θ = φ
@Irtsak
Ай бұрын
*No need of Trigonometry* Triangles ABD,ABC are similar => AB/x=3x/AB => AB²=3x²=>AB=x√3 Let apply Pythagoras theorem in right triangle ABD => AD²=AB²+BD² => AD²=3x²+x² => AD=2x Notice that AD=DC=2x => Δ ADC is isosceles =>
Another solution .................... Triangles ABD,ABC are similar => AB/x=3x/AB => AB²=3x²=>AB=x√3 Let apply Pythagoras theorem in right triangle ABC => AC²=AB²+BC² => AC²=3x²+9x² => AC=2√3 x Notice that AC=2√3 x =2(x√3)=2AB In right triangle ABC : AC=2AB => θ=30° Finish !
@daakudaddy5453
Ай бұрын
Unnecessarily complicated solution (first one, eapecially). From first look, you can say that ABD is similar to CBA (by AA, 90 degrees and theta common). Therefore, x/AB = AB/3x => AB = root 3. x Now, in ABD, as tan(theta) = x/root3.x = 1/root 3, Thus, theta = 30 degrees. Solved in 2 mins.
@Center1ForMathematics
Ай бұрын
This is an excellent work! there is another simple method that can be use to solve the given question. See solution to similar question here: kzitem.info/news/bejne/tXmwsoN5m3ZiZ6w
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