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Nice! One teaching point might be that only the slopes matter. Convert equations into y=mx+b form. One line has m=0.5 and the other m=3 Answer is tan-1(3)-tan-1(.5)=0.785 or 45 degrees. The difference in slopes gives the between angle Your way is a bit longer but it is very clear. Non-calculator approach for tests….use angle difference identities. Tan (a-b)= (tan(a)-tan(b))/(1 +tan(a)tan(b))= (3-.5)/(1+(3)(.5)) =1 therefore angle =45 degrees using unit circle chart.
this is for angles of a,b .. we want to do the opposite use inverse Tangent or aTan/ArcTan/Tan⁻¹ otherwise good .. y¹=k¹x+m¹ y²=k²x+m² -> angle θ between lines, θ = Tan⁻¹(k¹) - Tan⁻¹(k²)
Thank you very much for your kind feedback. I will be more careful from now on and take your valuable suggestions into consideration. I learn a lot from such encouraging comments. Best regards.
Hey friends! If you’re enjoying this video, could you double-check that you’ve liked it and suscribed to the channel? Thanks for helping me keep this going - you’re the best…
Emeğinize sağlık hocam...
Teşekkür ederim bro
Nice!
One teaching point might be that only the slopes matter.
Convert equations into y=mx+b form. One line has m=0.5 and the other m=3
Answer is tan-1(3)-tan-1(.5)=0.785 or 45 degrees.
The difference in slopes gives the between angle
Your way is a bit longer but it is very clear.
Non-calculator approach for tests….use angle difference identities.
Tan (a-b)= (tan(a)-tan(b))/(1 +tan(a)tan(b))= (3-.5)/(1+(3)(.5)) =1
therefore angle =45 degrees using unit circle chart.
this is for angles of a,b .. we want to do the opposite use inverse Tangent or aTan/ArcTan/Tan⁻¹
otherwise good ..
y¹=k¹x+m¹
y²=k²x+m²
->
angle θ between lines, θ = Tan⁻¹(k¹) - Tan⁻¹(k²)
Thank you very much for your kind feedback. I will be more careful from now on and take your valuable suggestions into consideration. I learn a lot from such encouraging comments. Best regards.
@@Patrik6920 Thank you very much for your kind comment.
Nice! 2y - x = 4 → y = x/2 + 2 = f(x); y - 3x = -5 → y = 3x - 5 = g(x) →
m1 = df(x)/dx = 1/2; m2 = dg(x)/dx = 3 → tan(θ) = (m1 - m2)/(1 + m1m2) = 1 → θ = 45°
or: φ = 30°; m2 = tan(α) = 3 → sin(α) = 3√10/10 → cos(α) = √10/10
m2 = tan(β) = 1/2 → sin(β) = √5/5 → cos(β) = 2√5)/5 →
sin(α - β) = sin(α)cos(β) - sin(β)cos(α) = (3√10/10)( (2√5)/5) - (2√5/5)(√10/10) = √2/2 →
α - β = 3φ/2
or: φ = 30°; tan(α) = 3 → cot(α) = tan(γ) = 1/3
tan(β) = 1/2 → arctan(γ) + arctan(β) = arctan( 3φ/2) → α - β = 3φ/2
Thank you very much for your nice comment