[3 marks sum]

Question 513 Marks

ABCD is a cyclic quadrilateral in a circle with centre O. If ∠ADC = 130°; find ∠ BAC.

Answer

Here ∠ACB = 90°
(Angle in a semicircle is right angle)
Also, ∠ABC = 180° -∠ADC = 180° - 130° = 50°
(pair of opposite angles in a cy clic quadrilateral are supplementary)
By angle sum property of right triangle ACB,
∠BAC = 90° - ∠ABC = 90° - 50° = 40°

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Question 523 Marks

In the following figure, O is centre of the circle and ΔABC is equilateral.
Find:(i) ∠ADB, (ii) ∠AEB.

Answer

Since ∠ACBand ∠ADBare in the same segment,
∠ADB = ∠ACB = 60°
Join OA and OB
Here, ∠AOB =2 ∠ACB = 2×60° = 120°
$\angle A E B=\frac{1}{2} \operatorname{Reflex}(\angle AOB )=\frac{1}{2}\left(360^{\circ}-120^{\circ}\right) 120^{\circ}$
(Angle at the centre is double the angle at the circumference subtended by the same chord)

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Question 533 Marks

In the figure, given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75°; ∠ABD = 58°
and ∠ADC = 77°. Find:
1. ∠BDC
2. ∠BCD
3. ∠BCA.

Answer

(i) By angle – sum property of triangle ABD,
∠BAD + ∠ABD + ∠ADB = 180°
133° + ∠ADB = 180°
∠ADB = 180° - 133°
∠ADB = 47°
∴ ∠ADC = ∠ADB + ∠BDC
∴ 77° = 47° + ∠BDC
∴ 77° - 47° = ∠BDC
∴ ∠BDC = 30°
(ii) ∠BAD + ∠BCD = 180° ...(Sum of opposite angles of a cyclic quadrilateral is 180°)
⇒ ∠BCD = 180° - 75° = 105°
∴ ∠BCD = 105°
(iii) ∠BCA = ∠BDA = 47° ...(Angle subtended by the same chord on the circle are equal)

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Mathematics [3 marks sum]