Geometry question

Question

Geometry question

mathmath333 · Answer

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mathmath333 · Answer

prove that 


$\large \color{black}{\begin{align} \dfrac{QS}{SR}=\dfrac{PQ}{PR}\hspace{.33em}\~\
\end{align}}$

mathmath333 · Answer

Given  that 
$\large \color{black}{\begin{align} \angle QPS=\angle RPS\hspace{.33em}\~\
\end{align}}$

rational · Answer

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rational · Answer

In $\triangle PQS$ applying sine law we get
$$\large \dfrac{QS}{PQ} = \dfrac{\sin \circ}{\sin x}\tag{1}$$

In $\triangle PRS$ applying sine law we get
$$\large \dfrac{SR}{PR} = \dfrac{\sin \circ}{\sin(180-x)}\tag{2}$$

rational · Answer

the right hand sides of $(1)$ and $(2)$ are same because $\sin(180-x) = \sin x$
equating left hand sides gives the desired proportion

mathmath333 · Answer

but y wolfram gives this identity wrong $\sin (180-x)=\sin x$ http://www.wolframalpha.com/input/?i=is+sind%28180-x%29%3Dsind+x+%3F

turingtest · Answer

it doesn't object to the same assertion in radians http://www.wolframalpha.com/input/?i=is+sin%28pi-x%29%3Dsin+x+%3F

mathmath333 · Answer

ok thanks