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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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Do u knw cos rule?
good:) so vat is cosA?

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Other answers:

b^2 + c^2 - a^2 / 2bc
a^2 +c^2 - b^2 / 2ac
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
2c^3 a^2 - 2a^4 c - 2c^3 b^2 + 2b^4 c = 0
u can factor out 2c.
2c ( b^4 - c^2 b^2 + a^2 c^2 - a^4) = 0 this right so far?
yup:) good:) nw divide by 2c both sides.
ok done that. btw im writing it down
u will get this. b^4 - c^2 b^2 + a^2 c^2 - a^4=0 try to factorise this.
(b^2 - a^2)(b^2 +a^2 -c^2)
(b+a)(b-a)(b^2+a^2 - c^2)
(b+a)(b-a)(b^2+a^2 - c^2)=0 Divide by (b+a)(b-a) both sides. vat do u get?
b^2 +a^2 - c^2 = 0
you can tell its right angle by looking at this.. but how do you tell if its isosceles?
u r nt asked to prove both right? either one of that?
oh i guess :)
i still think it means to prove that the triangle must be right or isoscles but it cant be scalene
but can we show it isoscles anyways?
let me think. hav to go nw. will be back in 15 mnts
(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.
ohh xDDDD THANKS!! Shame i can only give medal to 1 person :(
hey, i am not here for medals ;)
Welcome:) you can give it to @gohangoku58
cos he/she deserves it most:)
:O You made him solve the whole thing and i deserve it!??
yup:) it was u who pointed out the isoscles one:)

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