At 12:15, BD=2. Then ABC and ABD are similar because AB/BC = 1/3*sqrt(6) and BD/AB = 2/sqrt(6) = 1/3*sqrt(6), and ABC and ABD share angle ABC. Then angle ACB = angle BAD = 45 deg.
With BE perpendicular to AD, ∆AEB is Isosceles Right triangle, with BE=√3, ∠EBD=30° and BD =2. Reflect ∆BAD about AD to form ∆AFD. Since BF=2√3 and BC=3 and ∠FDC =60°, ∆FCD is a Right Traingle with ∠FCD =90°. Since ∠BAF=90°, BADC is a Cyclic Quadrilateral, with BF as its diameter which is also the base of the Right Triangle BAF in the circle with center at E. Since Chord BA subtends 90° at E, it subtends 45° at C. Hence θ= 45°
Thank you.. While solving a problem~ Two line segments AE and EC are equal ✓ 3 E=150° △ AEC is an isosceles triangle, so each ECA is 15° Therefore angle ACD=45° (두 선분 AE와 EC가 동일하게 ✓ 3 E=150° △ AEC는 이등변삼각형이니 각 ECA는 15° 그러므로 각ACD=45°)
As ∆ABD is a triangle, ∠ABD = 180°-(60°+45°) = 75°. By the law of sines: BD/sin(45°) = AB/sin(60°) BD = ABsin(45°)/sin(60°) BD = (√6/√2)/(√3/2) = (√6/√2)(2/√3) = 2 DA/sin(75°) = BD/sin(45°) DA = 2sin(45°+30°)/sin(45°) DA = 2(sin(45°)cos(30°)+cos(45°)sin(30°))/sin(45°) DA = 2(cos(30°)+sin(30°)) DA = 2(√3/2+1/2) = √3 + 1 Drop a perpendicular from A to E on BD. As ∠EDA = 60° and ∠AED = 90°, ∠DAE = 30°. In triangle ∆AED, ED = DAsin(30°) = (√3+1)/2. AE = DAcos(30°) = √3(√3+1)/2. In triangle ∆AEC, tan(θ) = AE/EC. tan(θ) = AE/EC = (√3(√3+1)/2)/((√3+1)/2)+1) tan(θ) = ((3+√3)/2)/((√3+1+2)/2) = 1 θ = tan⁻¹(1) = 45°
A little bit faster: After determining the first few lengths up to AE and ED, I just dropped the altitude from A onto BC at point M, and noticed that AM = CM so ACM is an isosceles right triangle.
AD = √6*sin 75/sin 60 AD= 2.732 AD /sin theta= 1 / sin(60-theta) Replace AD by 2.732 and solve the above equation for theta you will get Tan theta= 1 , which means theta= 45°
The answer is 45 degrees. At the 16:30 mark, I have noticed that you simplified cost15 degrees and sin15 degrees into (sqrt(3)+1)/(2sqrt(2)) and sqrt(3)-1)/(2sqrt(2)). Did you use the addition and subtraction identities of sine and cosine??? I want to know because I think that I am almost there in keeping track of every clever substitution, trig and algebraic required to be good at geometry. In fact, I have put down a list of potential problems to practice on in order to be good at those problems. Also sin15 degrees is the same as sin(90-15)=sin(75) and cos15=cos(15+75). Correct???
Both methods are too complex. Draw the perpendicular from B ot AD at E. Both AEB and DEB are special triangles - isosceles right and 30-60-90 triangles, respectively. As AD is given, easy to fund BE = AE = root 3. As BE = root 3, ED = 1. (Just applying dimensions of special triangles). Now, in triangle ADC, angle CAD = 60 - Theta (opposite angles theorem). Now apply sin rule in triangle ADC Sin theta / (1 + root 3) = sin ( 60 - Theta) / 1 Apply formula of Sin (A - B) to sin (60 - Theta), Simplify, We get Sin Theta = Cos Theta Thus, Theta = 45. Very easy thus way.
![Quando Rondo - Grow Up [Official Music Video]](http://i.ytimg.com/vi/8zFnCg3BO4Q/mqdefault.jpg)
At 12:15, BD=2. Then ABC and ABD are similar because AB/BC = 1/3*sqrt(6) and BD/AB = 2/sqrt(6) = 1/3*sqrt(6), and ABC and ABD share angle ABC. Then angle ACB = angle BAD = 45 deg.
Drop a perpendicular from A to BC and label the intersection as point E. Note that ΔADE is 30°-60°-90° special right triangle, so
With BE perpendicular to AD, ∆AEB is Isosceles Right triangle, with BE=√3, ∠EBD=30° and BD =2. Reflect ∆BAD about AD to form ∆AFD. Since BF=2√3 and BC=3 and ∠FDC =60°, ∆FCD is a Right Traingle with ∠FCD =90°.
Since ∠BAF=90°, BADC is a Cyclic Quadrilateral, with BF as its diameter which is also the base of the Right Triangle BAF in the circle with center at E.
Since Chord BA subtends 90° at E, it subtends 45° at C.
Hence θ= 45°
a really beautiful problem! Solved the same way.
Thank you..
While solving a problem~
Two line segments AE and EC are equal ✓ 3 E=150° △ AEC is an isosceles triangle, so each ECA is 15° Therefore angle ACD=45°
(두 선분 AE와 EC가 동일하게 ✓ 3
E=150°
△ AEC는 이등변삼각형이니 각 ECA는 15°
그러므로 각ACD=45°)
As ∆ABD is a triangle, ∠ABD = 180°-(60°+45°) = 75°. By the law of sines:
BD/sin(45°) = AB/sin(60°)
BD = ABsin(45°)/sin(60°)
BD = (√6/√2)/(√3/2) = (√6/√2)(2/√3) = 2
DA/sin(75°) = BD/sin(45°)
DA = 2sin(45°+30°)/sin(45°)
DA = 2(sin(45°)cos(30°)+cos(45°)sin(30°))/sin(45°)
DA = 2(cos(30°)+sin(30°))
DA = 2(√3/2+1/2) = √3 + 1
Drop a perpendicular from A to E on BD. As ∠EDA = 60° and ∠AED = 90°, ∠DAE = 30°.
In triangle ∆AED, ED = DAsin(30°) = (√3+1)/2. AE = DAcos(30°) = √3(√3+1)/2.
In triangle ∆AEC, tan(θ) = AE/EC.
tan(θ) = AE/EC = (√3(√3+1)/2)/((√3+1)/2)+1)
tan(θ) = ((3+√3)/2)/((√3+1+2)/2) = 1
θ = tan⁻¹(1) = 45°
φ = 30° → sin(3φ) = 1; sin(φ) = 1/2 → cos(φ) = √3/2 →
sin(5φ/2) = cos(φ/2) = √((1/2)(1 + cos(φ))) = (√2/4)(√3 + 1)
∆ ABC → AB = √6; BC = BD + CD = a + 1; AC = k; AD = b; DAB = 3φ/2
BDA = 2φ → ABD = 5φ/2 → ADC = 6φ - 2φ = 4φ → cos(4φ) = -cos(6φ - 4φ) = -1/2
∆ ABD → sin(2φ)/√6 = √2/4 = sin(3φ/2)/a = √2/2a → a = 2 → BC = 3 →
k^2 = 6 + 9 - 2(√6)3cos(5φ/2) = 3(2 + √3) → k = (√6/2)(√3 + 1) →
sin(θ)/√6 = sin(5φ/2)/k → sin(θ) = (√6/k)sin(5φ/2) = √2/2 → arcsin(√2/2) = θ = 3φ/2
A little bit faster: After determining the first few lengths up to AE and ED, I just dropped the altitude from A onto BC at point M, and noticed that AM = CM so ACM is an isosceles right triangle.
AD = √6*sin 75/sin 60
AD= 2.732
AD /sin theta= 1 / sin(60-theta)
Replace AD by 2.732 and solve the above equation for theta you will get
Tan theta= 1 , which means theta= 45°
(6)^2 =36;(1)^2= 1 {36+1}=37 {45°A+60°B}=105°AB {105°AB ➖ 75°C}= 30°ABC 30°ABC/37 5^6ABC/37^1 5^3^2ABC/1^1 5^13^2ABC/ 1^1^3^2ABC/ 3^2ABC (ABC ➖ 3ABC+2).
The answer is 45 degrees. At the 16:30 mark, I have noticed that you simplified cost15 degrees and sin15 degrees into (sqrt(3)+1)/(2sqrt(2)) and sqrt(3)-1)/(2sqrt(2)). Did you use the addition and subtraction identities of sine and cosine??? I want to know because I think that I am almost there in keeping track of every clever substitution, trig and algebraic required to be good at geometry. In fact, I have put down a list of potential problems to practice on in order to be good at those problems. Also sin15 degrees is the same as sin(90-15)=sin(75) and cos15=cos(15+75). Correct???
cos15° = cos(45° - 30°) , use cos(A-B) formula. By the same way, you can find sin15°.
sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
sin(a-b) = sin(a)cos(b) - cos(a)sin(b)
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
>> cos(a-b) = cos(a)cos(b) + sin(a)sin(b)
Never calculate Side lengths in a special triangle: learn it by heart
Col teorema dei seni risulta BD=2...ctgθ=(3/√6-cos75)/sin75=1...θ=45
45°
Both methods are too complex.
Draw the perpendicular from B ot AD at E. Both AEB and DEB are special triangles - isosceles right and 30-60-90 triangles, respectively.
As AD is given, easy to fund BE = AE = root 3.
As BE = root 3, ED = 1.
(Just applying dimensions of special triangles).
Now, in triangle ADC, angle CAD = 60 - Theta (opposite angles theorem).
Now apply sin rule in triangle ADC
Sin theta / (1 + root 3) = sin ( 60 - Theta) / 1
Apply formula of Sin (A - B) to sin (60 - Theta),
Simplify,
We get Sin Theta = Cos Theta
Thus, Theta = 45.
Very easy thus way.
45 degrees