SSC CGL 10 Jun 2019 Shift 2 Paper

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Question : 55 of 100

Marks: +1, -0

In a circle with centre

0, A B

is a diameter and

C D

is chord such that

\ {ABCD},

is a trapezium. If

\∠ {BAC}=15°

then

\∠ {CAD}

is equal to:

30°
60°
75°
45°

Solution:

\∠ {ACB}=90°

\∠ {CBA}+∠ {ADC}=180°

(opposite angles in a cyclic quadrilateral)

75°+∠ {ADC}=180°

∠ {ADC}=105°

∠ {BAC}=∠ {DCA}=15°(

alternate angle

)

\ ∴ ∠ {CAD}=180°-∠ {ADC}-∠ {DCA}

⇒∠ C A D=60°

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