Cosine law

**mido22** · September 6th 2011, 12:19 PM

Q : if $Cos B / a = Cos A / b$
Prove that : $a=b$ and $c^2$ $=$ $a^2$ $+$ $b^2$

**Opalg** · September 6th 2011, 11:27 PM

Originally Posted by mido22

Q : if $Cos B / a = Cos A / b$
Prove that : $a=b$ and $c^2$ $=$ $a^2$ $+$ $b^2$

What have you tried? You could start by using the cosine rule to express cos A and cos B in terms of a, b and c. Then try to simplify the resulting equation.

Edit. After checking the question, I think that you have stated it wrongly. It should say

Prove that : $a=b$ or $c^2=a^2+b^2$ .

**mido22** · September 7th 2011, 03:08 PM

yes it is or not and i made alot of tries but all reach to same result :
$b^2$ $($ $c^2$ $-$ $b^2$ $) =$ $a^2$ $($ $c^2$ $-$ $a^2$ $)$

**pickslides** · September 7th 2011, 03:25 PM

Originally Posted by mido22

yes it is or not and i made alot of tries but all reach to same result :
$b^2$ $($ $c^2$ $-$ $b^2$ $) =$ $a^2$ $($ $c^2$ $-$ $a^2$ $)$

Can you say a=b from this?

**mido22** · September 7th 2011, 03:35 PM

no i can't say so becz a!=b

**skeeter** · September 7th 2011, 04:22 PM

Originally Posted by mido22

yes it is or not and i made alot of tries but all reach to same result :
$b^2$ $($ $c^2$ $-$ $b^2$ $) =$ $a^2$ $($ $c^2$ $-$ $a^2$ $)$

$b^2(c^2-b^2) = a^2(c^2-a^2)$

$b^2c^2 - b^4 = a^2c^2 - a^4$

$b^2c^2 - a^2c^2 = b^4 - a^4$

$c^2(b^2 - a^2) = (b^2 - a^2)(b^2 + a^2)$

since $a \ne b$ ...

$c^2 = b^2 + a^2$

**mido22** · September 7th 2011, 04:25 PM

and how can i get a=b from same problem

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