th angle ADB = 65 degrees ( from the given data) and the triangle AED isosceles with two equal 65 degrees angles. The third is 50. Triangle AEC is also isosceles with the at a base angle=50. So, theta is 50 degrees.
E es punto medio de BD---> BE=EA=X ---> EABº=25º---> AEBº=180-25-25=130º---> AECº=180-130=50º ---> EA es simétrico de CA respecto a la vertical por A---> AECº=ACEº=50º. Gracias y saludos.
As ∠ABD = 25° and ∠DAB = 90°, ∠BDA = 180°-(90°+25°) = 65°. Let E be the midpoint of BD. Therefore BE = ED = x. Draw EA. By Thales' Theorem, as angle ∠DAB = 90°, if triangle ∆DAB were inscribed in a circle, BD would be the diameter and A would be a pint on the circumference, so EA = BE = ED = x. As BE = EA, ∆BEA is an isosceles triangle, so ∠EAB = ∠ABE = 25°. As ∠AED is an exterior angle to ∆BEA at E, ∠AED = ∠EAB+∠ABE = 25°+25° = 50°. As EA = AC = x, ∆EAC is an isosceles triangle, so ∠ECA = ∠AEC = θ = 50°.
if we create a circle surrounding ABD we get BD is diameter of circle. put a point in center of BD, let say it is point O so BO=DO=AO=x so AOD and AOC are isosceles triangles angle ADB = 180-90-25 = 65 from here angle ADO=ADB which is common, = DAO as isosceles = 65 so angle AOD=180-65-65 = 50 AOC is isosceles with angle AOC is 50 and AO=AC=x so angle ACO = AOC=50 but I vote for your 2nd method is cool
*_Outro método:_* Lei do seno no ∆ABC: sen 25°/x = sen θ/AB Como ∆ABD é retângulo, então, AB= 2x cos 25°, assim: sen 25°/x = sen θ/2x cos 25° sen θ = 2x sen 25° cos 25°/x sen θ = 2 sen 25° cos 25° como sen 2A = 2 sen A cos A, temos: sen θ = sen 50°, logo: θ= 50° ou θ=130°. No caso θ= 130°: O ângulo ∠ADC=90° + 25° = 115°. Consequentemente, A soma dos ângulos internos do ∆ACD ultrapassa 180°, absurdo! Portanto, *θ= 50°.*
I used the same method. It is a variant of the 2nd method with elimination of construction of altitude AE hence it is better. Your exclusion of 130 is likely to be missed by most students, me included.
th angle ADB = 65 degrees ( from the given data) and the triangle AED isosceles with two equal 65 degrees angles. The third is 50. Triangle AEC is also isosceles with the at a base angle=50. So, theta is 50 degrees.
E es punto medio de BD---> BE=EA=X ---> EABº=25º---> AEBº=180-25-25=130º---> AECº=180-130=50º ---> EA es simétrico de CA respecto a la vertical por A---> AECº=ACEº=50º.
Gracias y saludos.
First method is just so good!
I liked BOTH methods. And the angle theta is 50°. Then again I am familiar with BOTH methods btw.
As ∠ABD = 25° and ∠DAB = 90°, ∠BDA = 180°-(90°+25°) = 65°. Let E be the midpoint of BD. Therefore BE = ED = x. Draw EA. By Thales' Theorem, as angle ∠DAB = 90°, if triangle ∆DAB were inscribed in a circle, BD would be the diameter and A would be a pint on the circumference, so EA = BE = ED = x.
As BE = EA, ∆BEA is an isosceles triangle, so ∠EAB = ∠ABE = 25°. As ∠AED is an exterior angle to ∆BEA at E, ∠AED = ∠EAB+∠ABE = 25°+25° = 50°.
As EA = AC = x, ∆EAC is an isosceles triangle, so ∠ECA = ∠AEC = θ = 50°.
if we create a circle surrounding ABD we get BD is diameter of circle.
put a point in center of BD, let say it is point O so BO=DO=AO=x so AOD and AOC are isosceles triangles
angle ADB = 180-90-25 = 65
from here angle ADO=ADB which is common, = DAO as isosceles = 65 so angle AOD=180-65-65 = 50
AOC is isosceles with angle AOC is 50 and AO=AC=x so angle ACO = AOC=50
but I vote for your 2nd method is cool
{25°A+25°B+90°C+}=140°ABC {140°ABC ➖ 180x}=40°ABCX 2^20 2^2^20 1^1^2^10 2^2^5 1^2^1 2^1 (ABCX ➖ 2ABCX+1) .
Essa questão foi bem fácil. Parabéns pela escolha 👏👏👏👏👏👏👏👏 Brasil Outubro de 2024.
*_Outro método:_*
Lei do seno no ∆ABC:
sen 25°/x = sen θ/AB
Como ∆ABD é retângulo, então,
AB= 2x cos 25°, assim:
sen 25°/x = sen θ/2x cos 25°
sen θ = 2x sen 25° cos 25°/x
sen θ = 2 sen 25° cos 25°
como sen 2A = 2 sen A cos A, temos:
sen θ = sen 50°, logo:
θ= 50° ou θ=130°.
No caso θ= 130°:
O ângulo ∠ADC=90° + 25° = 115°. Consequentemente,
A soma dos ângulos internos do ∆ACD ultrapassa 180°, absurdo!
Portanto, *θ= 50°.*
I used the same method. It is a variant of the 2nd method with elimination of construction of altitude AE hence it is better. Your exclusion of 130 is likely to be missed by most students, me included.
@@hongningsuen1348 o bom é que sempre possamos contribuir para uma melhor compreensão da matemática! Abraços
Bravo❤
asnwer=45 isit
asnwer=50 isit
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@@ritwikgupta3655eu sei disso! Mas, não é melhor conferir antes de editar ou postar? Bom senso seria isso! Abraços
Teorema dei seni ADC..2xsin25/sinθ=x/sin115..sinθ=2sin25sin115=2sin25cos25=sin50...θ=50