Question

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc

Answer 1

Answer:

yaaa wait but the chord is not equal to radius of circle

Answer 2

$\huge\underlined{\underline{\sf {Given:}}}$

A circle with chord AB

AB = Radius of circle

Let point C be minor arc

& point D be mijor arc

$\huge\underlined{\underline{\sf {To Find:}}}$

Angle subtended by chord at a point in the minor arc, i.e., ∠ACB

& Angle subtended by a point in the major arc, i.e., ∠ADB.

$\huge\underlined{\underline{\sf{Construction:}}}$

Join OA & OB

$\huge\underlined{\underline{\sf{Explanation:}}}$

In ΔOAB ,

AB = OA = OB = radius

∴ ΔOAB is an equilateral triangle.

⇒  ∠AOB = 60° (All angles of equilateral triangle is 60°)

Arc  ADB makes ∠ AOB at centre

& angle ∠ADB at a point D

(Angle subtended by arc at a centre is  double the angle subtended by it at any other point)

So, ∠AOB = 2∠ADB

60° = 2∠ADB

2∠ADB = 60°

∠ADB = $\frac{1}{2}$ × (60°) = 30°

Also,

ADBC forms a cyclic quadrilateral

So,  ∠ADB + ∠ACB = 180° ( Sum of opposite angles of a cyclic quadrilateral                                                 is 180°)

30° + ∠ACB = 180°

∠ACB = 180° - 30°

∠ACB = 150°