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Prove that triangle is either an isosceles or a right angled triangle.

Posted August 4th, 2009 by alpha
in

If in a triangle $ \frac{a^2 -b^2 }{a^2 +b^2 }=\frac{sin(A-B)}{sin(A+B)} $

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answer

On August 6th, 2009 Structure says:
$$ \frac{a^2 -b^2 }{a^2 +b^2 }=\frac{sin(A-B)}{sin(A+B)}=\frac{\sin A \cos B-\sin B\cos A}{\sin A \cos B+\sin B\cos A} =\frac {a\frac{a^2+c^2-b^2}{2ac}-b\frac{b^2+c^2-a^2}{2bc}}{a\frac{a^2+c^2-b^2}{2ac}+b\frac{b^2+c^2-a^2}{2bc}}=\frac{a^2-b^2}{c^2}$$

So you have

$$a^2=b^2$$

or

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

How is a^2 equal to b^2?

On August 7th, 2009 alpha says:

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On August 8th, 2009 Structure says:

You can write

$$ \frac{a^2 -b^2 }{a^2 +b^2 }=\frac{sin(A-B)}{sin(A+B)}=\frac{\sin A \cos B-\sin B\cos A}{\sin A \cos B+\sin B\cos A} =\frac {a\frac{a^2+c^2-b^2}{2ac}-b\frac{b^2+c^2-a^2}{2bc}}{a\frac{a^2+c^2-b^2}{2ac}+b\frac{b^2+c^2-a^2}{2bc}}=\frac{a^2-b^2}{c^2}$$
$$ \frac{a^2 -b^2 }{a^2 +b^2 }=\frac{a^2-b^2}{c^2}$$
$$ \frac{a^2 -b^2 }{a^2 +b^2 }-\frac{a^2-b^2}{c^2}=0$$
$$ \frac{(a^2 -b^2 )(c^2-a^2-b^2)}{c^2(a^2 +b^2) }=0$$

~~THANK YOU ~~

On August 9th, 2009 alpha says:

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

Question on A.P, G.P, H.P

On September 1st, 2009 alpha says:

If a, b, c are in A.P , b,c,d, are in G.P, c,d,e are in H.P : prove that a,c,e are in G.P.

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

Permutation

On September 3rd, 2009 alpha says:

Prove that $ 1!+2 . 2!+ 3.3!+4.4!................. + n.n!=P(n+1,n+1)-1 $

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On September 5th, 2009 Structure says:

(n+1)!=(n+1)n!=n n!+n!
This is enough to solve your problem by induction

Permutation

On September 3rd, 2009 alpha says:

Prove that $ 1!+2 . 2!+ 3.3!+4.4!................. + n.n!=P(n+1,n+1)-1 $

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

Binomial Theorem:-

On September 12th, 2009 alpha says:

Find the remainder when 5^99 is divided by 13.

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On September 12th, 2009 Structure says:
$$5^{99}=5*5^{98}=5*25^{49}=5*(2*13-1)^{49}$$

As you can see the result is 8=13-5

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