Construct a pair of tangents to a circle which are inclined to each other at an angle of 60 degree
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Question ::
Construct a pair of tangents to a circle which are inclined to each other at an angle of 60 degree.
SOLUTION :::
We know that radius of the circle is perpendicular to the tangents.
Sum of all the 4 angles of quadrilateral = 360°
∴ Angle between the radius (∠O) = 360° - (90° + 90° + 60°) = 120°
✍Steps of Construction:✒
Step I: A point Q is taken on the circumference of the circle and OQ is joined. OQ is radius of the circle.
Step II: Draw another radius OR making an angle equal to 120° with the previous one.
Step III: A point P is taken outside the circle. QP and PR are joined which is perpendicular OQ and OR.
Thus, QP and PR are the required tangents inclined to each other at an angle of 60°.
✍Justification:✒
Sum of all angles in the quadrilateral PQOR = 360°
∠QOR + ∠ORP + ∠OQR + ∠RPQ = 360°
⇒ 120° + 90° + 90° + ∠RPQ = 360°
⇒∠RPQ = 360° - 300°
⇒∠RPQ = 60°
Hence, QP and PR are tangents inclined to each other at an angle of 60°.
Construction of a pair of tangents to a circle which are inclined to each other at an angle of 60° is in the attachment below.
Concept used:
We know that tangent is perpendicular to the radius at the point of contact.
∴ ∠D = ∠E = 90⁰
Given:
Angle between two tangents is 60⁰
∴ ∠C = 60⁰
We know that the Sum of all angles of a quadrilateral is 360⁰.
For quadrilateral ADCE,
∠A + ∠E + ∠C + ∠D = 360⁰
∠A + 90⁰ + 60⁰ + 90⁰ = 360⁰
∠A + 240⁰ = 360⁰
∠A = 120⁰
Steps for construction:
Step 1: Draw a circle with centre A and let radius be 3 cm
Step 2: Draw a radius AD and draw an angle of 120⁰ from radius AE. i.e ∠EAD = 120°.
Step 3: Draw perpendiculars EE’ and DD’ from points E and D.
Step 4: Extend the perpendiculars that intersect each other at point C. Then, CD and EC are the required tangents to the given circle inclined to each other at an angle of 60°.
Hence, the required tangents to the given circle inclined to each other at an angle of 60° is CD and CE.
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