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Maths TNSB Mentoring
Class 10 Crash course
Geometry
Annual Revision VI
3.
TBQ - Pythagoras theorem
Question:
1
m.
In the given figure, \(PR = 17 \ cm\), \(QR = 8 \ cm\), \(\angle PAQ = 90^\circ\), \(PA = 9 \ cm\) and \(QA = 12 \ cm\).
Find
\(\angle PQR\).
\(75^\circ\)
\(90^\circ\)
\(65^{\circ}\)
\(85^\circ\)
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