A Very Nice Geometry Problem | You should be able to solve this! | 2 Different Methods
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- เผยแพร่เมื่อ 24 ส.ค. 2024
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I have a third method . Extend line segment BA 2 units to point D . Next , draw a line from D to point C . You then have equilateral triangle BDC with three sides equal to x +2 and three inside angles of 60 degrees . Angle ADC is 60 degrees and angle DCB is 60 degrees . Now , consider triangle ADC . Use the law of sines on this triangle . We have angle ADC is 60 degrees and its opposite side is sqrt(6) . Call angle DCA alpha . Alpha's opposite side is 2 . We then have sin60/(sqrt(6))=sin(alpha)/2 . We can solve this equation to get sin(alpha)=1/(sqrt(2)) . This means that alpha=45 degrees . Now , alpha + theta =60 degrees where alpha =45 degrees . Therefore theta = 15 degrees . With this method , we don't need to calculate x .
great simple solution!
@@urban_sculptor Thanks . First , I solved the problem the hard way ; then , I thought of this simple method .
@@pk2712 even slightly more elegant - draw a line from A down to the base to create an internal equilateral triangle.
@@urban_sculptor I like the solution you are talking about also . I like math boosters solution . That why I have subscribed to this channel . Math Booster's solutions are some of the best detailed solutions on youtube . I always try to solve the problems that he posts before I watch his videos . Watching these videos is very enlightening .
Legge del coseno 6=x^2+(x+2)^2-2x(x+2)cos60...x=√3-1...legge del seno (√3-1)/sinθ=√6/sin60..sin θ=(√3-1)/2√2...θ=15
I solved it: 1st method - cosine rule: (√6) ^2 = (x+2) ^2 + x^2 - 2*(x+2) *x*cos (60) -> 2x^2 + 4x + 4 - x^2 - 2x = 6 --> x^2 + 2x - 2 = 0 -> x^2 - 2x + 1 = 3 --> x = √3-1. Then Sine Rule: x/sin (θ) = √6/sin (60) --> sin (θ) = x*sin (60) / √6 = (√3-1) * √3/ [2√6] * √6/√6 = [3√6 - 3√2]/12 = [√6 - √2]/4 --> θ = 15 Now, 2nd method is the first method you used and obtained identical answer.
The second method is preferable, as the sin & cos of 15 degrees (pi / 12) are not standard identities.
15
Draw a perpendicular from the vertex to the base to form a 90-60-30 right triangle on the left and another right triangle
on the right
Since "X" corresponds to the 90 degrees, "X/2" corresponds to the 30 degrees.
and "X/2 sqrt 3" corresponds to the 60 degrees
Let's employ the Pythagorean Theorem to find X using the right triangle on the right.
The shortest leg of the triangle to the right = X/2 sqrt 3
The longest leg of the triange to the right = (X+2 - X/2) = 2 + 0.5 X
The hypotenuse of the triangle to the right = sqrt 6
Hence,
(X/2 sqrt 3)^2 + (2 + 0.5 X) = (sqrt 6)^2
0.75 X^2 + 0.25X^2 + 2X + 4 =6
X^2 + 2 X - 2 =0. Use the quadratic equation to find X
X = 0.732051 OR SQRT 3 -1
Hence, theta = Sin 60 * 0.732051/ sqrt 6
theta = 15 degrees Answer
Find x by law of cosines:
6 = x² + (x+2)²- 2 . x . (x+2) . cos(60)
So: x = √ 3 - 1
Replace :
( √ 3 -1)² = (√ 3 +1)² + 6 - 2 .
(√ 3 + 1).√ 6. cos(θ)
So θ = 15°
The Laws of Cosines and Sines come in handy hefe
At 10:05, Math Booster jumps to the conclusion that if sin(θ) = (√3 - 1)/(2√2), then θ = 15°. This result is straightforward to derive from a knowledge of the side ratios for a 15°-75°-90° right triangle. They are (short):(long):(hypotenuse) (√3 - 1):(√3 + 1):2√2. sin(15°) = (short)/(hypotenuse) = (√3 - 1)/2√2, which matches the value of sin(θ) found in the video.
Note that sin(165°) = (√3 - 1)/(2√2) as well. However, if θ = 165°, the sum of angles in triangle ABC would exceed 180°, so θ = 165° is not a valid solution.
2nd method is awesome.Thank you professor.
Very useful to student community.
Thanks
Thank you for supporting the channel.
I like very much you teaching method ! Usually you detail so much than i find a bit too long… there is have been startled to see you going from square root (2+8)/2 directly to square root(1+2) 😅
< c = 30°
Delta=15°
Use cosine formula
Once I got sqrt 3-1 , sqrt 3 + 1 (from X + 2), I knew I was dealing with a 75, 15, 90 degrees which are in the ratio
2 sqrt 2, sqrt 3+ 1 and sqrt 3-1
Golly this problem is nice becuae it is also a refresher!!! I hope that that sentiment proves that I definitely know how to do this!!!
Math hunter was a better name
15
By the law of cosines:
cos(60°) = (BC²+AB²-CA²)/(2•BC•AB)
1/2 = ((x+2)²+x²-(√6)²)/(2x(x+2))
1/2 = (x²+4x+4+x²-6)/(2x²+4x)
1/2 = (2x²+4x-2)/(2x²+4x)
1/2 = (x²+2x-1)/(x²+2x)
2x² + 4x - 2 = x² + 2x
x² + 2x - 2 = 0
x = [-(2)±√(2²-4(1)(-2))]/2(1)
x = -1 ± √(4+8)/2
x = -1 ± √3
x = √3 - 1 | x = -√3 - 1 ❌ x ≥ 0
cos(θ) = (BC²+CA²-AB²)/(2•BC•CA)
cos(θ) = ((x+2)²+(√6)²-x²)/(2√6(x+2))
cos(θ) = (x²+4x+4+6-x²)/(2√6(x+2))
cos(θ) = (4(√3-1)+10)/(2√6(√3+1))
cos(θ) = 2(2√3+3)/2(3√2+√6)
cos(θ) = (2√3+3)(3√2-√6)/(3√2+√6)(3√2-√6)
cos(θ) = (6√6-6√2+9√2-3√6)/(18-6)
cos(θ) = (3√6+3√2)/12 = (√6+√2)/4
cos(θ) = (√3+1)/2√2
cos(θ) = √3/2√2 + 1/2√2
cos(θ) = (1/√2)(√3/2) + (1/√2)(1/2)
cos(θ) = cos(45°)cos(30°) + sin(45°)sin(30°)
cos(θ) = cos(45°-30°) = cos(15°)
θ = 15°
Very nice method
(6)^2/A/O/Asino°= 36A/O/ASino° {2x+2x ➖ }A/O/BCoso° =4x^2A/O/BCoso° {36A/O/ASino°+4x^2A/O/BCoso°}=40x^2A/O/ASino°BCoso° {60°A+60°B}=120°AB {120°AB+80°C}=180°ABC 180°ABC/40x^2A/O/ASino°BCoso°= 40.20x^2A/O/ASino°BCoso° 2^20.5^4 2^5^4.5^4 2^1^2^21^2^2 1^11^1.1^1^2 1^2 (A/O/ASino°B Coso°ABCX ➖ 2A/O/ASino°BCoso°ABCx+1).
φ = 30° → sin(3φ) = 1 → sin(2φ) = cos(φ) = √3/2 → cos(2φ) = sin(φ) = 1/2
∆ ABC → AB = x; BD = x + 2; AD = √6; ABD = 2φ → BC = BD + CD; BDA = θ = ?
∆ ABD → AB = x → BD = x/2 → AD = h = x√3/2; sin(BDA) = 1
∆ ABC → 6 = x^2 + (x + 2)^2 - 2x(x + 2)cos(2φ) → x = √3 - 1→ x√3/2 = h = (√3/2)(√3 - 1) →
sin(θ) = h/√6 = (√2/4)(√3 - 1) = sin(φ/2) = √((1/2)(1 - cos(φ))) → θ = φ/2
or: sin(2φ)/√6 = √2/4 = sin(θ)/(√3 - 1) → sin(θ) = (√2/4)(√3 - 1) = sin(φ/2) → θ = φ/2
😊
Please stop calling math problems "Nice"
"NICE" is correct. Please see Cambridge Dictionary
Not nice… but very nice. 😊