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Since $C=90^\circ$

Then $A+B=90$

Take, for instance $A=B=45$

So the origin equation becomes 1

I basically tried inputting the values in every option and arrived at $\frac {r_1}{R}$

Where r is the circumradius and $r_1$ is the radius of the excircle on side $A$.

How do I prove it properly?

Kenta S
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Aditya
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    You can't just assume that $A=B=45^\circ$. Having said that, this question looks unsolvable to me $-$ the value of $1+\sin A-\sin B$ can range from $0$ (if $A=0^\circ$) to $2$ (if $A=90^\circ$). Are you sure that you haven't left something out? – TonyK Dec 09 '19 at 12:47
  • Well, we can, since it is satisfied for all values as long as C=90 – Aditya Dec 09 '19 at 12:54
  • The way you put it ("We can assume...") means to me that the value of $1+\sin A-\sin B$ remains the same, whatever the values of $A$ and $B$. Which of course is not true. But then I don't understand what the question is asking for. – TonyK Dec 09 '19 at 12:58
  • The ‘absolute’ will change, but answer is in terms of variables, so they adjust accordingly. Yes, I should have chosen more appropriate words. – Aditya Dec 09 '19 at 13:01
  • So what is the question asking for? – TonyK Dec 09 '19 at 13:06
  • Just the ratio $\frac {r_1}{R}$ – Aditya Dec 09 '19 at 13:12
  • Please post the full question then! (If this is already the full question, then it makes no sense.) – TonyK Dec 09 '19 at 13:28

1 Answers1

1

Hint:

$$S=\sin C+\sin B-\sin A=2\sin\dfrac{B+C}2\cos\dfrac{B-C}2-2\sin\dfrac A2\cos\dfrac A2$$

As $\dfrac{B+C}2=\dfrac\pi2-\dfrac A2$

$$S=2\cos\dfrac A2\left(\cos\dfrac{B-C}2-\cos\dfrac{B+C}2\right)=4\cos\dfrac A2\sin\dfrac B2\sin\dfrac C2$$

Now use this

Here $C=90^\circ$

lab bhattacharjee
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  • So for the final part, the numerator would be r when it is $4R\sin \frac A2 \sin \frac B2 \sin \frac C2$ . How did you eliminate the cos? – Aditya Dec 09 '19 at 12:59
  • @Aditya, Please have a look into http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html and http://mathworld.wolfram.com/WernerFormulas.html. Please pinpoint the confusion – lab bhattacharjee Dec 09 '19 at 13:04