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If $\sin^2 A= \sin^2 B+\sin^2 C$, then the triangle is :

$1$. Equilateral

$2$. Isosceles

$3$. Acute angled triangle

$4$. Right angled triangle.

My Attempt:

$$\sin^2 A=\sin^2 B+\sin^2 C$$ $$1-\cos^2 A=1-\cos^2 B+ 1-\cos^2 C$$ $$\cos^2 A -\cos^2 B - \cos^2 C= -1$$

How do I proceed further?

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By the sine theorem $\sin^2 A=\sin^2 B+\sin^2 C$ implies $a^2=b^2+c^2$, so the given triangle is a right triangle with $\widehat{BAC}=90^\circ$, by the (converse) Pythagorean theorem.

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Hint: by the law of sines $\sin A = \cfrac{a}{2 R}\,$ and similar for the others. Then by Pythagoras' $\;\cdots$

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  • $\begingroup$ Could you please elaborate? I didn't understand. $\endgroup$ – pi-π Apr 27 '17 at 1:08
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    $\begingroup$ @Maxwell Lookup the law of sines. Then substitute the above into the given relation, factor out the common denominator of $4 R^2\,$, and you get $a^2=b^2+c^2$ where $a,b,c$ are the lengths of the sides. That makes the triangle a right one by Pythagoras' theorem. $\endgroup$ – dxiv Apr 27 '17 at 1:09
  • $\begingroup$ Not by Pythagoras - by inverse Pythagoras. $\endgroup$ – Alex Apr 27 '17 at 1:16
  • $\begingroup$ @dxiv What is $R$? $\endgroup$ – mrnovice Apr 27 '17 at 1:16
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    $\begingroup$ @Alex You mean the converse of Pythagoras. Right, though it depends on what you consider to be the theorem. It's sometimes stated as $a^2=b^2+c^2$ iff the triangle is a right triangle. $\endgroup$ – dxiv Apr 27 '17 at 1:18
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$$\sin^2A-\sin^2B=\sin^2C$$ $$\sin(A+B)\sin(A-B)=\sin^2C$$ $$\sin C(\sin(A-B)-\sin C)=0$$ $$\sin C(2\cos(\frac{A+C-B}{2})(\sin (\frac{A-(C+B)}{2})=0$$

Further hint: $A+C=\pi-B$ and $B+C=\pi-A$

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  • $\begingroup$ I didn't get the amswer. Please elaborate.. $\endgroup$ – pi-π Apr 27 '17 at 1:35
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Apply the law of sines together with the given condition: $$ {a\over\sin A} = {b\over\sin B} = {c\over\sin C} , \quad \sin^2 A =\sin^2 B +\sin^2 C \quad\Rightarrow\quad a^2=b^2+c^2. $$ Therefore you have a right triangle by the converse of the Pythagorean theorem.

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