A) \( 55 ^{\circ} \) |
B) \( 65 ^{\circ} \) |
C) \( 75 ^{\circ} \) |
D) \( 45 ^{\circ} \) |
B) \( 65 ^{\circ} \) |
Given : \( \angle SPQ = 50 ^{\circ} \)
To find : \( \angle RSQ \)
\( \angle SPQ = 50 ^{\circ} => \angle QRS = 50 ^{\circ} \)
since QR = RS
\( => \angle RQS = \angle RSQ = x \)
Therefore, In Trisngle QRS, we have
\( x + x ^{\circ} + 50 ^{\circ} = 180 ^{\circ} \)
=> \( 2x = 130 ^{\circ} \)
=> \( x = 65 ^{\circ} \)
=> \( \angle RSQ = 65 ^{\circ} \)