1 Marks Question

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° and $\angle$ECD = 20° Find $\angle$BAC.

Answer

Given: $\angle$BEC = 130° and $\angle$ECD = 20°
$\angle$DEC = 180° - $\angle$BEC = 180° - 130° = 50° [Linear pair]
Now in $\triangle$DEC,
$\angle$DEC + $\angle$DCE + $\angle$EDC = 180° [Angle sum property]
$\Rightarrow$ 50° + 20° + $\angle$EDC = 180° $\Rightarrow$ $\angle$EDC = 110°
$\Rightarrow$ $\angle$BAC = $\angle$EDC = 110° [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° (because angle in a semicircle is always 90° )
Similarly, we have $\angle$ABC = 90°
So, $\angle$ABD+ $\angle$ABC = 90° + 90° = 180°
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° and $\angle$BAC = 45°, find $\angle$BCD.

Answer

Since angles in the same segment of a circle are equal.
$\therefore$ $\angle$CAD = $\angle$DBC = 55^o
$\therefore$ $\angle$DAB = $\angle$CAD + $\angle$BAC = 55^o+ 45^o = 100^o
But, $\angle$DAB + $\angle$BCD = 180^o [Opposite angles of a cyclic quadrilateal]
$\therefore$ $\angle$BCD = 180^o - 100^o = 80^o

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