arctan(1/5) plus arctan(2) = 74.744... I found it easier to get the equation of each linear function and put them on an x-y graph.
@angeluomo
Күн бұрын
This was also my approach. I figured it was simpler, but it was interesting also to see the vector approach.
@deniseockey6204
Күн бұрын
Yes, this was not a good one. I would have done it like you did. If he wants to do vectors, he should use a better example.
@PreMath
11 сағат бұрын
Thanks for sharing ❤️
@danmike2305
4 сағат бұрын
Khan Academy also has vectors videos. Good review.
@wackojacko3962
Күн бұрын
This is a fun and intense problem. Somewhere down the line I'm thinkin it's a myth that vectors can't have negative magnitudes. I'll get back to ya on that. 🙂
This is nice, many thanks, Sir! φ = 30° → sin(3φ) = 1; f(x) = -(1/2)(x - 7) → x = 0 → f(x) = BO = 7/2 → f(x) = 0 → x = AO = 7 g(x) = 5x - 2 → x = 0 → g(x) = -2 = SO → g(x) = 0 → x = TO = 2/5 f(x) = g(x) → x = 13/11 → y = 43/11 → QO = y = 43/11; CO = x = 13/11 QO = PO = 43/11; sin(BQP) = 1 ∆ BQO → QPB = α → tan(α) = df(x)/dx = -1/2 → sin(α) = √5/5 → cos(α) =(2√5)/5 ∆ PQS → SPQ = β → tan(β) = dg(x)/dx = 5 → sin(β) = 5√26/26 → cos(β) = √26/26 6φ - (α + β) = θ = ? sin(γ) = sin(α + β) = sin(α)cos(β) + sin(β)cos(α) = 11√130/130 → cos(γ) = 3√130/130 sin(5φ/2) = (√2/4)(√3 + 1) > sin(γ) → sin(5φ/2 - γ) = sin(5φ/2)cos(γ) - sin(γ)cos(5φ/2) = (2√2)(√130)(7 - 4√3)/4(130) = (√260)(7 - 4√3)/260 = 0,004452646 ∶= k → (k/π)180° = 0,25511786° → 5φ/2 - (k/π)180° = 74,74488214° = γ → sin(γ) = sin(6φ - γ) = sin(θ) → α + β > 3φ → θ < 3φ → θ ≈ 74,745° or: φ = 30° → sin(3φ) = 1; circle around C → EC = BC = AC = r = √5 EQ = EP + PC + QC = EP + 2 + t; sin(BPC) = sin(CQA) = 1 AQ = AN + QN = (5t - 1) + 1; BP = QN = 1 sin(BNA) = 1; AB = n; PCB = α; ACQ = β ∆ ACQ → QC = t; AQ = 5t; AC = √5; sin(CQA) = 1 → t = √130/26 tan(α) = 1/2 → cot(α) = 2; tan(β) = 5 → cot(β) = 1/5; BCA = θ = ? arctan(θ) = arccot(α) + arccot(β) ≈ 74,745° or: ∆ ABN → BN = 2 + t; AN = (5t - 1); AB = n → n = √((5t - 1)^2 + (2 + t)^2) ∆ ACQ → QC = t; AQ = 5t; AC = r = √5; sin(CQA) = 1 → t = √130/26 → n^2 = 2(5)cos(θ) → cos(θ) = (1/10)(10 - n^2) = 3√130/130 → θ ≈ 74,745° or: tan(α) = -1/2; tan(β) = 5 → tan(θ) = |(tan(α) - tan(β))/(1 + tan(α)tan(β))| = 11/3 → arctan(θ) = 74,745°
@PreMath
11 сағат бұрын
Excellent! Thanks for sharing ❤️
@johanhaelterman853
Күн бұрын
Simpler to take the normal vector of the two equations as they have the same angle as the two lines: (5,-1) and (1,2) … the same formula for Theta can then be used. No need to calculate t. Nice exercise by the way ! J.
@PreMath
11 сағат бұрын
Thanks for the feedback ❤️
@vcvartak7111
Күн бұрын
Why not coordinate geometry formula tan(theta)=|m1-m2|/1+m1m2 where m1 and m2 are slope of lines which are -1/2 and 5 respectively
@PreMath
11 сағат бұрын
Thanks for the feedback ❤️
@robertlynch7520
Күн бұрын
I dunno much about vectors, so the trig-of-lines approach seemed to work out ok. Rearranging the line formulae to Ya = 5x - 2 and. Yb = 3.5 - 0.5x Then setting them equal finds crossing point of (x = 1, Ya=Yb = 3) From there, adding [ 1 ] to x gives Ya = 8 and Yb = 2.5 The ∆x = 1, the ∆Ya = 5 and ∆Yb = -0.5 The (not desired) big angle is [phi = alpha + beta] = [arctan(5) + arctan(0.5)] = 105.26 degrees. The desired angle [theta = 180 - phi] = 74.74 degrees, as the supplement angle of a line. Ta Da. GoatGuy
@PreMath
11 сағат бұрын
Excellent! Thanks for the nice feedback ❤️
@marcgriselhubert3915
Күн бұрын
VectorU(1; 5) is director of the straigt line whose equation is y = 5.x -2, and VectorV(-2; 1) of the other one. (More generally if the equation is a.x + b.y +c = 0, then VectorW(-b; a) is director. Then = -2+5 = 3 = ||VectorU||.||VectorV||.cos(theta) = sqrt(26).sqrt(5).cos(theta), so cos(theta) = 3/sqrt(130) and theta = cos-1(3/sqrt(130).
@unknownidentity2846
Күн бұрын
Let's find the angle: . .. ... .... ..... First of all we calculate the point of intersection: 5x − y = 2 x + 2y = 7 10x − 2y = 4 x + 2y = 7 11x = 11 ⇒ x = 1 ⇒ y = 5x − 2 = 5*1 − 2 = 5 − 2 = 3 ∧ y = (7 − x)/2 = (7 − 1)/2 = 6/2 = 3 ✓ Other points on the blue and red line are for example: x = 2 ⇒ y = 5x − 2 = 5*2 − 2 = 10 − 2 = 8 x = −1 ⇒ y = (7 − x)/2 = (7 + 1)/2 = 8/2 = 4 The angle θ is the angle between these two vectors: v(b) = ( 2 − 1 ; 8 − 3 ) = ( 1 ; 5 ) v(r) = ( −1 − 1 ; 4 − 3 ) = ( −2 ; 1 ) Now we are able to calculate the angle θ: cos(θ) = v(b) * v(r) / (|v(b)| * |v(r)|) = [1*(−2) + 5*1] / {√(1² + 5²) * √[(−2)² + 1²]} = (−2 + 5) / [√(1 + 25) * √(4 + 1)] = 3 / (√26 * √5) = 3 / √130 ⇒ θ = arccos(3 / √130) ≈ 74.74° Best regards from Germany
@PreMath
11 сағат бұрын
Excellent! Thanks for sharing ❤️
@Birol731
Күн бұрын
My way of solution ▶ Let's check the first line: 5x-y=2 y= 5x-2 dy/dx= 5 tan(α)= 5 α= arctan(5) α= 78,69° the second line: x+2y= 7 2y= 7-x y= (7-x)/2 dy/dx= -1/2 β= arctan(-1/2) β= -26,5650° tan(-β)= -tan(β) ⇒ β= 26,5650° The angle Θ, Θ= 180°-(78,69°+26,5650°) Θ= 74,745° Θ≈ 74,75°
L1 : 5x - y = 2 => y = 5x - 2 L2 : x + 2y = 7 => y = (-1/2)x + 7/2 angle between L1 and X-axis = Θ1, sin(Θ1) = 5/✓26 angle between L2 and X-axis = Θ2, sin(Θ2) = (1/2)/✓(5/4) = 1/✓5 sin(Θ1 + Θ2) = sin(Θ1)cos(Θ2) + cos(Θ1)sin(Θ2) = (5/✓26)(2/✓5) + (1/✓26)(1/✓5) = 11/✓130 Θ = π - (Θ1 + Θ2) = π - arcsin(11/✓130)
@PreMath
11 сағат бұрын
Thanks for sharing ❤️
@calvinmasters6159
Күн бұрын
Cartesian method rather than vectors. Observe x-coefficients as slopes: y = 5x - 2 y = -1/2(x) + 7 atn(1/5) + atn(2) ≈ 74.74 deg
@DareChristopher
Күн бұрын
y = -x/2 + 7/2
@calvinmasters6159
Күн бұрын
@@DareChristopher Ah, you're right. Clumsy me. But, since we're not using the y-intercept, it doesn't matter for the answer.
@egillandersson1780
Күн бұрын
I get the same with the formula of tan (β-𝜶) where tan 𝜶 = 5 and tan β = -1/2 => tan (β-𝜶) = 11/3 and arctan 11/3 ≃ 74,74°
@PreMath
11 сағат бұрын
Thanks for sharing ❤️
@jimlocke9320
21 сағат бұрын
The video presents the vector approach. I'll present a different approach. First, rewrite the equations in the form y = f(x): Blue line y = 5x - 2, red line y = -x/2 + 7/2. Set the right sides equal to each other and solve: 5x - 2 = -x/2 + 7/2, 11x/2 = 11/2, x = 1. Substitute x = 1 into either y = equation and find y = 3. So, the coordinates of the intersection are (1 , 3). Construct a line upward from the intersection. Its equation is x = 1. Let the acute angle between the red line and the portion of the vertical line from the intersection upward, which is a ray, be designated α and between the blue line and the ray be ß. Construct a horizontal line distance m above where the red and blue lines intersect. It has equation y = m + 3 and its intersection with the vertical line has coordinates (1 , m + 3). It intersects the red line at (1 - 2m, m + 3) and the blue line at (1 + m/5, m + 3). Two right triangles are formed, one with a red hypotenuse and the other a blue hypotenuse. The side opposite angle α has length 2m and opposite angle ß has length m/5. The other side in both triangles is common and has length m. tan(α) = 2m/m = 2 and tan(ß) = (m/5)/m = 1/5. tan(Θ) = tan(α + ß) = (tan(α) + tan(ß))/(1 - (tan(α))(tan{ß))) = (2 + 1/5)/(1 - (2)(1/5)) = (11/5)/(1 - 2/5) = 11/3. So Θ = arctan(11/3) or approximately 74.74°. If we construct a right triangle with sides 11 and 3, the hypotenuse has length √(130) and angle Θ is opposite the side of length 11, so cos(Θ) = 3/(√(130)) and Θ = arccos(3/(√(130))), as found in the video.
@PreMath
11 сағат бұрын
Excellent! Thanks for sharing ❤️
@sujoymukherjee1930
Күн бұрын
How can both x and y be equal to t?
@michaeldoerr5810
Күн бұрын
The angle theta is 74.74°. I take that as a refresher for practicing the vector approach.
@PreMath
11 сағат бұрын
Excellent! Glad to hear that! Thanks for the feedback ❤️
@michaelgarrow3239
Күн бұрын
There is no coronation between the length of the lines and the angle they intersect.
@PreMath
11 сағат бұрын
Thanks for the feedback ❤️
@LuisdeBritoCamacho
Күн бұрын
STEP-BY-STEP RESOLUTION PROPOSAL : 01) Red Line, Plane Equation : X + 2Y = 7 02) Blue Lin, Plane Equation : 5X - Y = 2 03) Normal Vector of Red Line : n(R) = (1 ; 2) 04) Normal Vector of Blue Line : n(B) = (5 ; -1) 05) Director Vector of Red Line : d(R) = (-2 ; 1) or (2 ; -1) 06) Director Vector of Blue Line : d(B) = (1 ; -5) or (-1 ; 5) 07) Dot Product between the two Vectors : DP = (1 ; 2) * (5 ; -1) ; DP = (1 * 5 ) + (2 * -1) ; DP = 5 + (-2) ; DP = 3 08) So, as DP is different from 0 (DP = 3); the two Lines are not Perpendicular to each other. 09) Finding Theta Angle : 10) Theta = arc cos | n(R) * n(B) / (|| n(R) || * || n(B||) | 11) || n(R) || = sqrt(5) 12) || n(B) || = sqrt(26) 13) Theta = arc cos | 3 / sqrt(130) | 14) Theta = arc cos ( ~ 0,263) 15) Theta ~ 74,745º Thus, OUR BEST ANSWER : The Tetha Acute Angle equal approx. 74,745º.
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