In the given figure, A B C D E is a pentagon inscribed in a circle such that A C is a diameter and side B C ∥ A E . If ∠ B A C = 50 ° , find giving reason ∠ A C B .

We have, In the given figure, A B C D E is a pentagon inscribed in a circle such that A C is a diameter and side B C ∥ A E . If ∠ B A C = 50 ° , Now, ∠ A B C = 90 ° (Angle in a semicircle) In ∆ A B C , ∠ A B C + ∠ B A C + ∠ A C B = 180 ° (angle sum property of a triangle) 90 ° + 50 ° + ∠ A C B = 180 ° 140 ° + ∠ A C B = 180 ° ∠ A C B = 180 ° - 140 ° ∠ A C B = 40 °

In the given figure, A B C D E is a pentagon inscribed in a circle such that A C is a diameter and side B C ∥ A E . If ∠ B A C = 50 ° , find giving reason ∠ E D C .

Given, Since A C is a diameter and angle in a semi-circle is right angle ∠ B = 90 ° And, ∠ A C B = 180 ° - ∠ B - ∠ B A C = 180 ° - 90 ° - 50 ° = 40 ° Also, B C ∥ A E ⇒ ∠ E A C = ∠ A C B = 40 ° [alternate interior angles] In cyclic quadrilateral A C D E , ∠ E A C + ∠ E D C = 180 ° [opposite angles are supplementary] ⇒ 40 ° + ∠ E D C = 180 ° ⇒ ∠ E D C = 180 ° – 40 ° = 140 ° Hence, ∠ E D C = 140 °

In the given figure, A B C D E is a pentagon inscribed in a circle such that A C is a diameter and side B C ∥ A E . If ∠ B A C = 50 ° , find giving reason ∠ B E C

Given figure: Join B E and E C Given, ∠ B A C = 50 ° So, ∠ B E C and ∠ B A C as they are angles in same segment formed by chord B C . So, they are equal. ∠ B E C = 50 ° .

In the given figure A C is a tangent to the circle with centre O . If ∠ A D B = 55 ° , find x a n d y . Give reasons for your answers.

Given, We know that angle between the radius and the tangent at the point of contact is the right angle. ∠ A = 90 ° Also, in ∆ O B E , O B = O E = r a d i u s r In Δ A B D , ∠ A + ∠ B + ∠ A D B = 180 ° 90 ° + ∠ B + 55 ° = 180 ° ∠ B = 180 ° − 90 ° − 55 ° = 35 ° ∠ B = ∠ O E B = 35 ° a n d ∠ O E B = ∠ D E C = 35 ° (vertically opposite angles) ∠ E D C + ∠ A D E = 180 ° ∠ E D C + 55 ° = 180 ° ∠ E D C = 180 ° − 55 ° = 125 ° I n Δ E D C , ∠ D E C + ∠ E D C + x = 180 ° 35 ° + 125 ° + x = 180 ° x = 180 ° − 35 ° − 125 ° = 20 ° I n Δ A O C , ∠ A + ∠ y + ∠ x = 180 0 90 ° + y + 20 ° = 180 ° y = 180 ° − 90 ° − 20 ° = 70 ° Hence, the value of x = 20 ° a n d y = 70 ° .

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 15 - Circles>ICSE-2020>Q 10

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In the figure given below, $O$ is the centre of the circle and $A B$ is a diameter. If $A C = B D$ and $∠ A O C = 72 °$ . Find $∠ A B D$ in degrees.

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Important Questions on Circles

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10th ICSE

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 15 - Circles>ICSE-2020>Q 11

In the given figure $T P$ and $T Q$ are two tangents to the circle with centre $O$ , touching at $A$ and $C$ respectively. If $∠ B C Q = 55 °$ and $∠ B A P = 60 °$ , find $∠ O B A$ and $∠ O B C$

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 15 - Circles>ICSE-2020>Q 12

In the given figure $T P$ and $T Q$ are two tangents to the circle with centre $O$ , touching at $A$ and $C$ respectively. If $∠ B C Q = 55 °$ and $∠ B A P = 60 °$ , find $∠ A O C .$

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10th ICSE

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 15 - Circles>ICSE-2020>Q 13

In the given figure $T P & T Q$ are two tangents to the circle with centre $O$ , touching at $A & C$ respectively. If $∠ B C Q = 55 ° & ∠ B A P = 60 °$ , find $∠ A T C$ in degrees.