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SSC CGL Model Paper 1

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Question : 52 of 100

Marks: +1, -0

In the given figure, a tangent TA touches the circle with centre 'O' at A. AC is a chord of the circle. ∠ABC =115°. Then find ∠CAT?

85°
65°
70°
60°

Solution:

\angle ABC + \angle ADC = 180^{\circ}

(opposite angles of cyclic quadrilateral)

\angle ADC = 180^{\circ} - 115^{\circ} = 65^{\circ}

\angle AOC = 2\angle CDA

[0 being circumcentre of

\triangle ADC

]

\angle AOC = 2 \times 65 = 130^{\circ}

Now, OA = OC [radii of the given circle]

So,

\angle OAC = \angle OCA

In

\triangle OAC; \angle OAC + \angle OCA + \angle AOC = 180^{\circ}

2\angle OAC = 180^{\circ} - 130^{\circ}

\angle OAC = 25^{\circ}

Also, OA is

\perp^{\tau}

to TA

So,

\angle CAT = \angle ZOAT - \angle OAC

= 90^{\circ} - 25^{\circ} = 65^{\circ}

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