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.
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.....
*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 !
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.
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: czcams.com/video/TCzLOCbD14w/video.html
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°
yes, I solved this the same method. Only the difference is in calculation of hypotenuse and then the sin of Theta.
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.
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.
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°
Short form
Vertical height y then
Tan@=x/y
Tan@=y/3x
Equating x/y is 1/√3
Therefore theta is 30°
x/AB = AB/3x
AB^2 = 3•x^2
AB = sqrt(3)•x
tanC = sqrt(3)•x / 3x = 1/sqrt(3)
Thus, C = 30degree
❤❤
🎉🎉❤❤🎉🎉
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
30 degrees!
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.....
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
*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 =>
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°
(x/sinθ):sinθ=2x:sin(90-2θ)...2(sinθ)^2=1-2(sinθ)^2..sinθ/=1/2..θ=30
(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)
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 !
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.
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: czcams.com/video/TCzLOCbD14w/video.html