AonZ
Help please :) Yr11 3 unit question on trig
In triangle ABC, acosA = bcosB. Prove using the cosine rule, that the triangle is either isosceles or right-angled
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ajprincess
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Do u knw cos rule?
AonZ
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yep...
ajprincess
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good:) so vat is cosA?
AonZ
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b^2 + c^2 - a^2 / 2bc
ajprincess
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cosB=?
AonZ
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a^2 +c^2 - b^2 / 2ac
ajprincess
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good:)
ajprincess
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Nw if u r to prove that the triangle is isoscles u need to show that a=b or b=c or a=c. if u r to show it is a right angled triangle u should show pythagoras theorem is valid.
\[a\cos A=b\cos B\]
\[a* \frac{ b^2+c^2-a^2 }{ 2bc}=b*\frac{ a^2+c^2-b^2 }{ 2ac }\]
vat do u get when u simplify this? @AonZ
AonZ
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2c^3 a^2 - 2a^4 c - 2c^3 b^2 + 2b^4 c = 0
ajprincess
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u can factor out 2c.
AonZ
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2c ( b^4 - c^2 b^2 + a^2 c^2 - a^4) = 0
this right so far?
ajprincess
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yup:) good:) nw divide by 2c both sides.
AonZ
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ok done that. btw im writing it down
ajprincess
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ok:)
ajprincess
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u will get this.
b^4 - c^2 b^2 + a^2 c^2 - a^4=0
try to factorise this.
AonZ
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(b^2 - a^2)(b^2 +a^2 -c^2)
AonZ
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(b+a)(b-a)(b^2+a^2 - c^2)
ajprincess
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(b+a)(b-a)(b^2+a^2 - c^2)=0
Divide by (b+a)(b-a) both sides. vat do u get?
AonZ
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b^2 +a^2 - c^2 = 0
AonZ
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you can tell its right angle by looking at this.. but how do you tell if its isosceles?
ajprincess
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u r nt asked to prove both right? either one of that?
AonZ
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oh i guess :)
AonZ
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i still think it means to prove that the triangle must be right or isoscles but it cant be scalene
AonZ
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but can we show it isoscles anyways?
ajprincess
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let me think. hav to go nw. will be back in 15 mnts
gohangoku58
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(b+a)(b-a)(b^2+a^2 - c^2) =0
gives
(b+a)=0 OR (b-a)= 0 OR (b^2+a^2 - c^2)=0
so, b-a = 0 --->a=b----->isosceles
(b^2+a^2 - c^2)=0 --->right as you already know.
AonZ
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ohh xDDDD THANKS!!
Shame i can only give medal to 1 person :(
gohangoku58
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hey, i am not here for medals ;)
ajprincess
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Welcome:) you can give it to @gohangoku58
ajprincess
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cos he/she deserves it most:)
gohangoku58
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:O
You made him solve the whole thing and i deserve it!??
AonZ
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lol
ajprincess
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yup:) it was u who pointed out the isoscles one:)