AB is the diameter of the circle-CD is parallel to AB - - CAB -32- Find - ADC-A- 110-B- 105-C- 122-D- 140-

The correct option is C 122^∘ ∠ ACB=∠ ADB=90^∘ (Angle subtended by the diameter is 90^o) In ABC ∠ B = 180 ( 90 + 32 ) = 58^o Therefore ∠ ADC = 180 58 = ...

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Standard X

Mathematics

Properties of Cyclic Quadrilateral

AB is the dia...

Question

AB is the diameter of the circle.CD is parallel to AB, $∠ C A B = 32^{\circ}$ . Find $∠ A D C$ .

$110^{\circ}$

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$105^{\circ}$

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$122^{\circ}$

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$140^{\circ}$

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Solution

The correct option is C
$122^{\circ}$

$∠ A C B = ∠ A D B = 90^{\circ}$ (Angle subtended by the diameter is $90^{o}$ )

$I n △ A B C ∠ B = 180 - (90 + 32) = 58^{o}$

Therefore $∠ A D C = 180 - 58 = 122^{o}$ (ABCD is a cyclic quadrilateral and opposite angles are supplementary)

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AB is the diameter of the circle.CD is parallel to AB,∠CAB=32∘. Find ∠ADC.

The correct option is C 122∘ ∠ACB=∠ADB=90∘ (Angle subtended by the diameter is 90o) In △ABC ∠B=180−(90+32) = 58o Therefore ∠ADC=180−58=122o (ABCD is a cyclic quadrilateral and opposite angles are supplementary)

AB is the diameter of the circle.CD is parallel to AB, $∠ C A B = 32^{\circ}$ . Find $∠ A D C$ .

The correct option is C
$122^{\circ}$

$∠ A C B = ∠ A D B = 90^{\circ}$ (Angle subtended by the diameter is $90^{o}$ )

$I n △ A B C ∠ B = 180 - (90 + 32) = 58^{o}$

Therefore $∠ A D C = 180 - 58 = 122^{o}$ (ABCD is a cyclic quadrilateral and opposite angles are supplementary)