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24 Jun 2021, 06:15
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (02:46) correct
63%
(02:36)
wrong
based on 63
sessions
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In the given figure, the two circles are concentric at centre O. SQ, SU are tangents to the smaller circle and QP is a tangent to the bigger circle. Also, SQ = SU. If ∠PQR = 80°, what is the measure of ∠RST ?
(A) 20°
(B) 40°
(C) 60°
(D) 80°
(E) 100°
Attachment:
2021-06-16_18-38-26.png [ 16.9 KiB | Viewed 2252 times ]
24 Aug 2021, 23:56
Kudos
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The solution to this is :-
join OQ and OS. OQ and OS are the radii of the bigger circle. Angle OQR is 10 degrees as the angle made by radius and tangent is 90 degrees and RQP is 80 degrees. Now that OQ and OS are equal angle OSR is 10 degrees. Multiply it by 2 gives us RST as 20 degrees
join OQ and OS. OQ and OS are the radii of the bigger circle. Angle OQR is 10 degrees as the angle made by radius and tangent is 90 degrees and RQP is 80 degrees. Now that OQ and OS are equal angle OSR is 10 degrees. Multiply it by 2 gives us RST as 20 degrees
25 Aug 2021, 07:45
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Bunuel
Connect OV, OU, and OS as shown in the graph below. We know the lines are tangent so ∠OVP is a right angle. Then ∠RVO is 10 degrees, using symmetry we can figure out ∠OUT is also 10 degrees. Then using isosceles triangle VOS and UOS, we can find ∠VSU = 10 + 10 = 20.
Ans: A
Attachments
circles.png [ 36.28 KiB | Viewed 1690 times ]
