A Very Nice Geometry Problem | You should be able to solve this! | 2 Methods

[@quigonkenny]

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]

yes, I solved this the same method. Only the difference is in calculation of hypotenuse and then the sin of Theta.

[@michaeldoerr5810]

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.

[@juanalfaro7522]

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.

[@Irtsak]

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°

[@kinno1837]

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°

[@cosmolbfu67]

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]

Short form Vertical height y then Tan@=x/y Tan@=y/3x Equating x/y is 1/√3 Therefore theta is 30°

[@ManojkantSamal]

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.....

[@Irtsak]

*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 =>

[@haiduy7627]

❤❤

[@RealQinnMalloryu4]

(2x)^2 =4x^2 3A(15°)= 45°A 3B(15°)=45°B {45°A+45°B+90°} 180°AB 180°AB/4x = 4x.20AB 2^2.5^4AB 1^1.1^4 2^2 (ABx ➖ 2ABx+2)

[@murdock5537]

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

[@haiduy7627]

🎉🎉❤❤🎉🎉

[@tituschirila7650]

as previous posters: x/AB=tan(Θ); AB/3x=tan(Θ); AB=x/tan(Θ); x/tan(Θ)*3x=tan(Θ); => tan(Θ)^2 = 1/3; tan(Θ)= sqrt(1/3); Θ = 30º

[@giuseppemalaguti435]

(x/sinθ):sinθ=2x:sin(90-2θ)...2(sinθ)^2=1-2(sinθ)^2..sinθ/=1/2..θ=30

[@Irtsak]

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 !

[@sergeyvinns931]

30 degrees!

[@daakudaddy5453]

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]

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: kzbin.info/www/bejne/inTdfYJ5l6lkatk