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In the diagram, ABCD is a square. Find sin PAQ.

 Feb 15, 2024

Best Answer 

 #1
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We know \(AD=BA=3+2=5\) 

\(AQ=AP= \sqrt{AD^2+DQ^2}=\sqrt{34}\)

\(\triangle AQD \cong \triangle ABP\) by SSS congruence
Thus:
\(\angle QAD=\angle PAB\) (by congruence above)
And \(\angle PAQ=90-2\angle PAB\) 
 Let \(\angle PAB =x\)and \(\angle PAQ=90-2x\)    (for simplicity)
 

\(\sin(90-2x)=\cos(2x)=1-2\sin^2x\)

We know \(\sin x=\frac {PB}{AP}=\frac{3}{\sqrt{34}}\)
so finally:

\(\sin {\angle PAQ}=\boxed{\frac{8}{17}}\)
ggwp

 Feb 15, 2024
 #1
avatar
+1
Best Answer

We know \(AD=BA=3+2=5\) 

\(AQ=AP= \sqrt{AD^2+DQ^2}=\sqrt{34}\)

\(\triangle AQD \cong \triangle ABP\) by SSS congruence
Thus:
\(\angle QAD=\angle PAB\) (by congruence above)
And \(\angle PAQ=90-2\angle PAB\) 
 Let \(\angle PAB =x\)and \(\angle PAQ=90-2x\)    (for simplicity)
 

\(\sin(90-2x)=\cos(2x)=1-2\sin^2x\)

We know \(\sin x=\frac {PB}{AP}=\frac{3}{\sqrt{34}}\)
so finally:

\(\sin {\angle PAQ}=\boxed{\frac{8}{17}}\)
ggwp

EnormousBighead Feb 15, 2024

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