"O" es punto medio de BD y centro de un semicírculo que pasa por A→ OA=OB=OD=AC→ El triángulo ACO es isósceles → Ángulos: ABO=25º=BAO→ AOB=130º→ AOC=50º=ACO=θ. Gracias y saludos.
Theta angle is 50 degrees. Here is my summary of the two methods involved: The first method as proven by circle theorems or without simply shows that corresponding sides and angles are congruent. The second method uses basic trig identities because of the definition of a perpendicular bisector. I think that even the perpendicular bisector also follows from corresponding sides and angles being congruent. I could be wrong.
50 degrees Answer Another approach Triangle ABD is a 25-65-90 degree triangle (given) Since BD = 2x, then the length AD = 2x ( sine 25/sine 90) degrees AD =0.84523652348 x Hence, AB = 1.81262 x Hence, the three sides of triangle ABC are 2x, 1.81262x, and 0.84523652348 in terms of x Draw a perpendicular from BD to A to form another 25-65-90 degree right triangle. APD (P is the new point) Hence, AP/0.845 = 1.81262/2 since triangle APD is similar to triangle ABD. Hence, AP = 0.845 * 1.81262/2 AP =0.7659925 x (recall it is in terms of x) Hence, the length PD = 0.35732(using the length of AP, AD, and Pythagorean Theorem). Will get back to using PD later Let's calculate PC using AP and AC or 0.7659925 x and x (given) PC = 0.64285 x (Pythagorean) My goal is to find the length of BC: Since BD = 2x (given) and PD =0.35732 x (see above) (see "Will get back to using PD later") Then BP = 1.64268x Since PC = 0.64285x (see above) Then BC = BP + PC= 1.64268x + 0.64285x BC =2.28553x Now remove all the inner triangles and focus on triangle ABC AC= x (given) or 1 x and BC = 2.28553x (see above) Using the Law of sines 1/sine 25 degrees= 2.28553/ angle A Angle A*1 = 2.28553 * sine 25 degrees (cross multiply) 2.35520158315 angle A = 2.28553/2.35520158315 Angle A = 2.28553x * sin 25 degrees)/x Angle A = 2.28553* sin 25 degrees/1 (remove the x ) Angle A = 0. 96590671575 A = 105 degrees as the sine105 degrees = 0.96590671575 Hence, Angle A = 105 degrees Since angle B= 25 degrees (given) Then angle C = 50 (180- (105 + 25) = 50 (180-130) Answer
OMG, I'm so thankful for this problem because my intuition has been lacking the last few problems. I'm in awe at how I saw this one so swiftly. My reasoning goes in line with your first solution except I assigned "x" a value of "1" : )
As ∆DAB is a triangle, ∠BDA = 180°-90°-25° = 65°. As ∠ADC is an exterior angle to ∆DAB at D, ∠ADC = 90°+25° = 115°. As ∠BDA is an exterior angle to ∆DCA at D, ∠CAD = 65°-θ. By the law of sines: AD/sin(25°) = BD/sin(90°) AD/sin(25°) = 2x/1 AD = 2xsin(25°) ---- [1] CA/sin(115°) = AD/sin(θ) x/sin(65°) = AD/sin(θ) x/cos(25°) = 2xsin(25°)/sin(θ)
Here is my version with trigonometry only to share with. sin 25=AD/2x -->AD=2x*sin 25. …..(1) Angle ADB=65 ,hence angle ADC=180-65 Applying the law of sine for triangle ADC sin(theta)/AD=sin(180-65)/x=sin 65/x …..(2) Replacing (1) to (2) sin(theta)=2*sin 25*sin 65=-cos 90+cos 40=cos 40=sin 50 Hence theta=50
At about 11:25, Math Booster finds that sin(Θ) = sin(50°) and concludes that Θ = 50°. Actually, there is a second solution, Θ = 130°, which satisfies sin(Θ) = sin(50°), from the identify sin(α) = sin(180° - α). From the diagram, it is clear that Θ is an acute angle, so 130° is not valid. However, the solution Θ = 130° really should be presented and then discarded as being not valid for this problem because the angle must be acute. A different problem might produce sin(Θ) = sin(50°) where Θ = 130° is the correct answer.
Second method's result so satisfying
"O" es punto medio de BD y centro de un semicírculo que pasa por A→ OA=OB=OD=AC→ El triángulo ACO es isósceles → Ángulos: ABO=25º=BAO→ AOB=130º→ AOC=50º=ACO=θ.
Gracias y saludos.
Theta angle is 50 degrees. Here is my summary of the two methods involved: The first method as proven by circle theorems or without simply shows that corresponding sides and angles are congruent. The second method uses basic trig identities because of the definition of a perpendicular bisector. I think that even the perpendicular bisector also follows from corresponding sides and angles being congruent. I could be wrong.
50 degrees Answer
Another approach
Triangle ABD is a 25-65-90 degree triangle (given)
Since BD = 2x,
then the length AD = 2x ( sine 25/sine 90) degrees
AD =0.84523652348 x
Hence, AB = 1.81262 x
Hence, the three sides of triangle ABC are 2x, 1.81262x, and 0.84523652348 in terms of x
Draw a perpendicular from BD to A to form another 25-65-90 degree right triangle. APD (P is the new point)
Hence, AP/0.845 = 1.81262/2 since triangle APD is similar to triangle ABD.
Hence, AP = 0.845 * 1.81262/2
AP =0.7659925 x (recall it is in terms of x)
Hence, the length PD = 0.35732(using the length of AP, AD, and Pythagorean Theorem). Will get back to using PD later
Let's calculate PC using AP and AC or 0.7659925 x and x (given)
PC = 0.64285 x (Pythagorean)
My goal is to find the length of BC:
Since BD = 2x (given)
and PD =0.35732 x (see above) (see "Will get back to using PD later")
Then BP = 1.64268x
Since PC = 0.64285x (see above)
Then BC = BP + PC= 1.64268x + 0.64285x
BC =2.28553x
Now remove all the inner triangles and focus on triangle ABC
AC= x (given) or 1 x
and BC = 2.28553x (see above)
Using the Law of sines
1/sine 25 degrees= 2.28553/ angle A
Angle A*1 = 2.28553 * sine 25 degrees (cross multiply)
2.35520158315
angle A = 2.28553/2.35520158315
Angle A = 2.28553x * sin 25 degrees)/x
Angle A = 2.28553* sin 25 degrees/1 (remove the x )
Angle A = 0. 96590671575
A = 105 degrees as the sine105 degrees = 0.96590671575
Hence, Angle A = 105 degrees
Since angle B= 25 degrees (given)
Then angle C = 50 (180- (105 + 25)
= 50 (180-130) Answer
OMG, I'm so thankful for this problem because my intuition has been lacking the last few problems. I'm in awe at how I saw this one so swiftly.
My reasoning goes in line with your first solution except I assigned "x" a value of "1" : )
As ∆DAB is a triangle, ∠BDA = 180°-90°-25° = 65°. As ∠ADC is an exterior angle to ∆DAB at D, ∠ADC = 90°+25° = 115°. As ∠BDA is an exterior angle to ∆DCA at D, ∠CAD = 65°-θ.
By the law of sines:
AD/sin(25°) = BD/sin(90°)
AD/sin(25°) = 2x/1
AD = 2xsin(25°) ---- [1]
CA/sin(115°) = AD/sin(θ)
x/sin(65°) = AD/sin(θ)
x/cos(25°) = 2xsin(25°)/sin(θ)
Here is my version with trigonometry only to share with.
sin 25=AD/2x -->AD=2x*sin 25. …..(1)
Angle ADB=65 ,hence angle ADC=180-65
Applying the law of sine for triangle ADC
sin(theta)/AD=sin(180-65)/x=sin 65/x …..(2)
Replacing (1) to (2)
sin(theta)=2*sin 25*sin 65=-cos 90+cos 40=cos 40=sin 50
Hence theta=50
(2x)^2^=4x^2 {25°A+25°B+90°C}=140ABC {140°ABC+40D}= 180°ABCD {180°ABCD/4x^2}=4.20ABCDx^2 2^2.5^4ABCDx^2 1^1.5^12^2ABCDx^2 1^11^1ABCDx^2 1ABCDx^2 (ABCDx ➖ 2ABCDx+1).
Second method is reaaly nice. Congrats
Thank you 🙂
At about 11:25, Math Booster finds that sin(Θ) = sin(50°) and concludes that Θ = 50°. Actually, there is a second solution, Θ = 130°, which satisfies sin(Θ) = sin(50°), from the identify sin(α) = sin(180° - α). From the diagram, it is clear that Θ is an acute angle, so 130° is not valid. However, the solution Θ = 130° really should be presented and then discarded as being not valid for this problem because the angle must be acute. A different problem might produce sin(Θ) = sin(50°) where Θ = 130° is the correct answer.
Theta is acute angle, so it will be 50°
@@MathBooster Agreed! I just think that the second solution should have been presented and then discarded, for the reason you give.
Esse segundo método foi lindo demais !
50 degrees
با رسم میانه A و BD زاویه تتا به. راحتی مشخص میشه
(2xsin25)/sinθ=x/sin115..sinθ=2sin115sin25=2cos25sin25=sin50
2(sin115)(sin25) is just a number, .766; arcsin .766 = 50° = θ
But your solution is - mathematically - more elegant🤓