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Two unequal circles with centres A and B...

Two unequal circles with centres A and B intersect each other at points C and D. The centre B of the smaller circles lies on the circumference of the bigger circle with centre A. If ` angle CMD = x^(@) , ` find in terms of x, the measure of angle DAC .
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To solve the problem, we need to find the measure of angle DAC in terms of x, given that angle CMD = x degrees. Here is the step-by-step solution: ### Step 1: Understand the Geometry We have two circles with centers A and B that intersect at points C and D. The center B of the smaller circle lies on the circumference of the larger circle centered at A. ### Step 2: Identify Tangents Since B lies on the circumference of the larger circle, the lines CB and BD are tangents to the larger circle at points C and D respectively. Therefore, angles CBM and BDN are right angles (90 degrees). ### Step 3: Relate Angles We know that angle CMD is given as x degrees. The angle subtended at the center A by the chord CD is twice the angle subtended at any point on the circumference (which is angle CMD). Therefore, we can write: \[ \angle CAB = 2x \] ### Step 4: Use the Sum of Angles in Quadrilateral In quadrilateral ACBD, the sum of the angles is equal to 360 degrees. The angles we have are: - Angle CAB = 2x - Angle CBM = 90 degrees - Angle BDN = 90 degrees - Angle CND (which we will express in terms of angle CAD) Thus, we can write the equation: \[ 90 + 90 + 2x + \angle CND = 360 \] ### Step 5: Solve for Angle CND From the equation above, we can simplify: \[ 180 + 2x + \angle CND = 360 \] Subtracting 180 and 2x from both sides gives: \[ \angle CND = 180 - 2x \] ### Step 6: Relate Angle CND to Angle CAD The angle CND is half of angle CAD (since it subtends the same arc CD). Therefore, we can write: \[ \angle CND = \frac{1}{2} \angle CAD \] Substituting the expression we found for angle CND: \[ 180 - 2x = \frac{1}{2} \angle CAD \] ### Step 7: Solve for Angle CAD To find angle CAD, we multiply both sides by 2: \[ 2(180 - 2x) = \angle CAD \] This simplifies to: \[ \angle CAD = 360 - 4x \] ### Step 8: Find Angle DAC Since angle DAC is the same as angle CAD, we can conclude that: \[ \angle DAC = 360 - 4x \] ### Final Answer Thus, the measure of angle DAC in terms of x is: \[ \angle DAC = 360 - 4x \] ---
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ICSE-CIRCLES-EXERCISE 17( C )
  1. Two unequal circles with centres A and B intersect each other at point...

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  2. In the given circle with diameter AB, find the value of x.

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  3. In the given figure , ABC is a triangle in which angle BAC = 30 ^(@) ...

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  4. Prove that the circle drawn on any one of the equal sides of an iso...

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  5. In the given figures, chord ED is parallel to diameter AC of the circl...

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  6. The quadrilateral formed by angle bisectors of a cyclic quadrilateral ...

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  7. In the figure angle DBC = 58^(@) , BD is a diameter of the circle . ...

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  8. D and E are points on equal sides A B and A C of an isosceles triangle...

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  9. In the given figure ,ABCD is a cyclic quadrilateral .AF is drawn para...

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  10. If I is the incentre of triangle ABC and AI when produced meets the c...

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  11. In the given figure ,AB = AD = DC = PB and angle DBC = x^(@) Determ...

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  12. In the given figure , ABC , AEQ and CEP are straight lines . Show that...

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  13. In the given figure. AB is the diameter of the circle with centre O. ...

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  14. In a cyclic -quadrilateral PQRS angle PQR = 135^(@) , Sides SP and...

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  15. In the following figure , ABCD is a cyclic quadrilateral in which AD i...

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  16. ABCD is a cyclic quadrilateral , Sides AB and DC produced meet at poin...

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  17. The following figure shows a circle with PR as its diameter. If PQ= ...

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  18. In the given figure AB is the diameter of a circle with centre O . If ...

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  19. In cyclic quadrilateral ABCD , AD = BC , angle BAC = 30 ^(@) and angl...

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  20. In cyclic quadrilateral ABCD , AD = BC , angle BAC = 30 ^(@) and angl...

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  21. In cyclic quadrilateral ABCD , AD = BC , angle BAC = 30 ^(@) and an...

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