A) \[30{}^\circ \]
B) \[15{}^\circ \]
C) \[45{}^\circ \]
D) \[60{}^\circ \]
Correct Answer: B
Solution :
PS = SQ |
\[\angle QRS=\frac{1}{2}[180{}^\circ -120{}^\circ ]=30{}^\circ \] |
\[\angle QRP=(180{}^\circ -30{}^\circ )=150{}^\circ \] |
Hence, \[\angle QPR=\frac{1}{2}(180{}^\circ -150{}^\circ )=15{}^\circ \] |
\[\because \](PR = RQ) |
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