# Show that ....

• Dec 15th 2010, 07:41 AM
razemsoft21
Show that ....
ABC is a triangle with sides a,b and c. show that :
if $a^2-b^2=bc$ then A=2B

I try to use the cosine law, but no result
• Dec 15th 2010, 06:37 PM
tonio
Quote:

Originally Posted by razemsoft21
ABC is a triangle with sides a,b and c. show that :
if $a^2-b^2=bc$ then A=2B

I try to use the cosine law, but no result

In any isosceles straight-angle triangle with hypotenuse $a$ and legs $b\,,\,c=b$ we have, by Pythagoras, that

$a^2=b^2+b^2=b^2+bc\Longrightarrow a^2-b^2=bc(=b^2)$ , but nevertheless $a=\sqrt{2}b\neq 2b$ ,

and thus the claim is false.

Tonio
• Dec 16th 2010, 05:09 AM
Soroban
Hello, razemsoft21!

Quote:

$ABC\text{ is a triangle with sides }a,b\text{ and }c.$

$\text{Show that: if }a^2-b^2\:=\:bc,\,\text{ then }A\,=\,2B.$

We are given: . $\begin{Bmatrix}a^2-b^2 &=& bc \\ a^2-bc &=& b^2 \\ b^2 + bc &=& a^2 \end{Bmatrix}$

Law of Cosines:

$\displaystyle \cos A \;=\;\frac{b^2+c^2-a^2}{2bc} \;=\;\frac{c^2-(a^2-b^2)}{2bc} \;=\;\frac{c^2-bc}{2bc} \;=\;\frac{c(c-b)}{2bc} \;=\;\frac{c-b}{2b}$ .[1]

$\displaystyle \cos B \;=\;\frac{a^2+c^2 - b^2}{2ac} \;=\;\frac{c^2 + (a^2-b^2)}{2ac} \;=\;\frac{c^2 + bc}{2ac} \;=\;\frac{c(c+b)}{2ac} \;=\;\frac{c+b}{2a}$

$\displaystyle \cos2B \;=\;2\cos^2\!B - 1 \;=\;2\left(\frac{c+b}{2a}\right)^2 - 1 \;=\;\frac{b^2 + 2bc + c^2 - 2a^2}{2a^2}$

. . . . . . $\displaystyle =\;\frac{2bc + c^2 - a^2 - \overbrace{(a^2-b^2)}^{\text{This is }bc}}{2\underbrace{(b^2+bc)}_{\text{This was }a^2}} \;=\;\frac{2bc + c^2 - a^2 - bc}{2b(b+c)}$

. . . . . . $\displaystyle =\;\frac{bc + c^2 - a^2}{2b(b+c)} \;=\;\frac{c^2 - \overbrace{(a^2 - bc)}^{\text{This is }b^2}}{2b(b+c)} \;=\;\frac{c^2 - b^2}{2b(b+c)}
$

. . . . . . $\displaystyle =\;\frac{(c-b)(c+b)}{2b(c+b)} \;=\;\frac{c-b}{2b}$ .[2]

Since [1] = [2], we have: . $\cos A \:=\:\cos2B$

. . Therefore: . $A \,=\,2B$

• Dec 16th 2010, 05:27 AM
razemsoft21
Quote:

Originally Posted by tonio
In any isosceles straight-angle triangle with hypotenuse $a$ and legs $b\,,\,c=b$ we have, by Pythagoras, that

$a^2=b^2+b^2=b^2+bc\Longrightarrow a^2-b^2=bc(=b^2)$ , but nevertheless $a=\sqrt{2}b\neq 2b$ ,

and thus the claim is false.

Tonio

Our triangle is not isosceles, and it is not aright-angled triangle.
and I think we cann't use Pythagoras theorem.
• Dec 16th 2010, 06:12 AM
tonio
Quote:

Originally Posted by Soroban
Hello, razemsoft21!

We are given: . $\begin{Bmatrix}a^2-b^2 &=& bc \\ a^2-bc &=& b^2 \\ b^2 + bc &=& a^2 \end{Bmatrix}$

Law of Cosines:

$\displaystyle \cos A \;=\;\frac{b^2+c^2-a^2}{2bc} \;=\;\frac{c^2-(a^2-b^2)}{2bc} \;=\;\frac{c^2-bc}{2bc} \;=\;\frac{c(c-b)}{2bc} \;=\;\frac{c-b}{2b}$ .[1]

$\displaystyle \cos B \;=\;\frac{a^2+c^2 - b^2}{2ac} \;=\;\frac{c^2 + (a^2-b^2)}{2ac} \;=\;\frac{c^2 + bc}{2ac} \;=\;\frac{c(c+b)}{2ac} \;=\;\frac{c+b}{2a}$

$\displaystyle \cos2B \;=\;2\cos^2\!B - 1 \;=\;2\left(\frac{c+b}{2a}\right)^2 - 1 \;=\;\frac{b^2 + 2bc + c^2 - 2a^2}{2a^2}$

. . . . . . $\displaystyle =\;\frac{2bc + c^2 - a^2 - \overbrace{(a^2-b^2)}^{\text{This is }bc}}{2\underbrace{(b^2+bc)}_{\text{This was }a^2}} \;=\;\frac{2bc + c^2 - a^2 - bc}{2b(b+c)}$

. . . . . . $\displaystyle =\;\frac{bc + c^2 - a^2}{2b(b+c)} \;=\;\frac{c^2 - \overbrace{(a^2 - bc)}^{\text{This is }b^2}}{2b(b+c)} \;=\;\frac{c^2 - b^2}{2b(b+c)}
$

. . . . . . $\displaystyle =\;\frac{(c-b)(c+b)}{2b(c+b)} \;=\;\frac{c-b}{2b}$ .[2]

Since [1] = [2], we have: . $\cos A \:=\:\cos2B$

. . Therefore: . $A \,=\,2B$

Very nice. In my example I confused between $a, A\,\,and\,\,b, B$ , since clearly $A=\frac{\pi}{2}=2\frac{\pi}{4}=2B$...*sigh*

Of course, this would have hardly happend had the OP written $\angle A=2\angle B$ ...

Tonio
• Dec 18th 2010, 04:36 PM
Quote:

Originally Posted by razemsoft21
ABC is a triangle with sides a,b and c. show that :

if $a^2-b^2=bc$ then A=2B

I tried to use the cosine law, but no result.

Another way....

$a^2-b^2=\left[b^2+c^2-2bcCosA\right]-\left[a^2+c^2-2acCosB\right]=b^2-a^2-2bcCosA+2acCosB$

$\Rightarrow\ 2\left(a^2-b^2\right)=2bc=2\left[acCosB-bcCosA\right]\Rightarrow\ b=aCosB-bCosA$

$\displaystyle\Rightarrow\ b(1+CosA)=aCosB\Rightarrow\frac{b}{a}=\frac{CosB}{ 1+CosA}$

We have an alternative expression for this fraction using the Sine Law

$\displaystyle\frac{a}{SinA}=\frac{b}{SinB}\Rightar row\frac{b}{a}=\frac{SinB}{SinA}$

$\displaystyle\Rightarrow\frac{CosB}{1+CosA}=\frac{ SinB}{SinA}\Rightarrow\ SinACosB=SinB+SinBCosA$

$\Rightarrow\ SinACosB-CosASinB=SinB$

Using the identity $Sin(A-B)=SinACosB-CosASinB$

$\Rightarrow\ Sin(A-B)=SinB\Rightarrow\ A=2B$
• Dec 19th 2010, 02:51 PM
razemsoft21
Quote:

Another way....

$a^2-b^2=\left[b^2+c^2-2bcCosA\right]-\left[a^2+c^2-2acCosB\right]=b^2-a^2-2bcCosA+2acCosB$

$\Rightarrow\ 2\left(a^2-b^2\right)=2bc=2\left[acCosB-bcCosA\right]\Rightarrow\ b=aCosB-bCosA$

$\displaystyle\Rightarrow\ b(1+CosA)=aCosB\Rightarrow\frac{b}{a}=\frac{CosB}{ 1+CosA}$

We have an alternative expression for this fraction using the Sine Law

$\displaystyle\frac{a}{SinA}=\frac{b}{SinB}\Rightar row\frac{b}{a}=\frac{SinB}{SinA}$

$\displaystyle\Rightarrow\frac{CosB}{1+CosA}=\frac{ SinB}{SinA}\Rightarrow\ SinACosB=SinB+SinBCosA$

$\Rightarrow\ SinACosB-CosASinB=SinB$

Using the identity $Sin(A-B)=SinACosB-CosASinB$

$\Rightarrow\ Sin(A-B)=SinB\Rightarrow\ A=2B$

GOOD job .... under stood .... THANKS