SSC CGL 11 Jun 2019 Shift 1 Paper

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Question : 69 of 100

Marks: +1, -0

In a circle with centre O, AB is the diameter and CD is a chord such that ΔBCD is a trapezium. If ∠BAC = 25°, then ∠CAD is equal to:

45°
65°
40°
25°

Solution:

\ {\ABCD} \→

Trapezium

A B ∥ C D \⇒\∠C A B=\∠A C D=25°

And

\∠ {\ACB}=90°

(angle in semi-circle)

\∵\∠A B C=180°-\ (25°+90°)=65°

And

\∠ {\CDA}=180°-65°=115°

Hence,

\∠ {\CAD}=180°-\ (115°+25°)=40°

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