In the given figure, ABCD is a cyclic quadrilateral whose side AB is a diameter of the circle passing through A,B,C and D. If ∠ADC =130o, find ∠BAC.

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A
$45^{o}$
B
$58^{o}$
C
$60^{o}$
D
$40^{o}$

Solution:

Consider the given figure,

Here,

∠

ADC

= 130^{o}

.
Angle subtended by the diameter of circle is

90^{o}

.
Then,

∠ ACB = 90^{o}

. Also, in a cyclic quadrilateral sum of opposite angles are supplementary.
Then in quadrilateral ABCD,

\Rightarrow ∠ ABC + ∠ ADC = 180^{0}

\Rightarrow ∠ ABC = 180 ° - ∠ ADC

\Rightarrow ∠ ABC = 180 ° - 130 °

\Rightarrow ∠ ABC = 50 °

In triangle

ABC

, by angle sum property,

\Rightarrow ∠ BAC + ∠ ACB + ∠ ABC = 180 °

\Rightarrow ∠ BAC = 180 ° - (∠ ACB + ∠ ABC)

\Rightarrow ∠ BAC = 180 ° - 50 ° - 90 °

\Rightarrow ∠ BAC = 180 ° - 140 °

\Rightarrow ∠ BAC = 40^{o}

Hence, the measure of

∠ BAC

40^{o}

.
Therefore, option 4 is correct.

In the given figure, $ABCD$ is a cyclic quadrilateral whose side $AB$ is a diameter of the circle passing through $A, B, C$ and $D$ . If $∠$ ADC $= 130^{o}$ , find $∠$ BAC.

Solution:

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