Question

how to draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

Answer 1

$\tiny \red{\bold{ hello \: dear }}$

# Step 1

Draw a circle of radius 5 cm and with centre as O.

# Step 2

Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A.

# Step 3

Draw a radius OB, making an angle of 120° (180° − 60°) with OA.

# Step 4

Draw a perpendicular to OB at point B. Let both the perpendiculars intersect at point P. PA and PB are the required tangents at an angle of 60°.

##### justification :-

The construction can be justified by proving that ∠APB = 60°

∠OAP = 90°

∠OBP = 90°

∠AOB = 120°

sum of all interior angles of a quadrilateral = 360°

∠OAP + ∠AOB + ∠OBP + ∠APB = 360°

90° + 120° + 90° + ∠APB = 360°

∠APB = 60°

Answer 2

hey dear !!


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here's your answer...



Question :: how to draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.






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.



$hope \: \\ this \: \\ helps \: u \: ....$