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Question 11 Mark

In figure, $A, B, C, D$ are four points on the circle. $AC$ and $BD$ intersect at a point $E$ such that $\angle BEC = 130^\circ $ and $\angle ECD = 20^\circ $ Find $\angle BAC.$

Answer

Given: $\angle BEC = 130^\circ $ and $\angle ECD = 20^\circ $
$ \angle DEC = 180^\circ - \angle BEC = 180^\circ - 130^\circ = 50^\circ [$Linear pair$]$
Now in $\triangle DEC,$
$ \angle DEC + \angle DCE + \angle EDC = 180^\circ [$Angle sum property$]$
$\Rightarrow 50^\circ + 20^\circ + \angle EDC = 180^\circ \Rightarrow \angle EDC = 110^\circ $
$\Rightarrow \angle BAC = \angle EDC = 110^\circ [$Angles in same segment$]$

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Question 21 Mark

In the given figure, two circles intersect at two points $A$ and $B. AD$ and $AC$ are diameters to the circles. Prove that $B$ lies on the line segment $DC.$

Answer

In the given diagram join $AB.$ Also $\angle ABD = 90^\circ ($because angle in a semicircle is always $90^\circ )$
Similarly, we have $\angle ABC = 90^\circ $
So, $\angle ABD+ \angle ABC = 90^\circ + 90^\circ = 180^\circ $
Therefore, $DBC$ is a line i.e., $B$ lies on the line segment $DC.$

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Question 31 Mark

In Fig., $ABCD$ is a cyclic quadrilateral in which $AC$ and $BD$ are its diagonals. If $\angle DBC = 55^\circ $ and $\angle BAC = 45^\circ ,$ find $\angle BCD.$

Answer

Since angles in the same segment of a circle are equal.
$\therefore \angle C A D=\angle D B C=55^{\circ}$
$\therefore \angle D A B=\angle C A D+\angle B A C=55^{\circ}+45^{\circ}=100^{\circ}$
But, $\angle \mathrm{DAB}+\angle \mathrm{BCD}=180^{\circ} [$Opposite angles of a cyclic quadrilateal$]$
$\therefore \angle B C D=180^{\circ}-100^{\circ}=80^{\circ}$

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