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Two chords of length a unit and b unit of a circle make angles 60° and 90° at the centre of a circle respectively, then the correct relation is (SSC CGL 1st Sit. 2015)
- a)b = √2 a
- b)b = 2a
- c)b = √3a
- d)
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Two chords of length a unit and b unit of a circle make angles 60°...
Let the chord AB = a and chord BC = b makes angle
∠ AOB = 60° and ∠ BOC = 90° at the center ‘O’ of the circle. There, OA = OB = OC = radius
In ΔAOB, ∠ OAB = ∠ OBA
and ∠ AOB = 60°
∴ ∠ OAB + ∠ OBA = 180° – 60° = 120°
⇒ ∠ OAB + ∠ OAB = 120°
⇒ ∠ OAB = 60°
Thus, ∠ OAB = ∠ OBA = ∠ AOB = 60°
∴ AOB is equilateral triangle
Hence, AO = OB = AB = a unit
Now, from ΔBOC, ∠ BOC = 90°, BC = b unit
OB = OC = a unit
(BC)2 = (OB)2 + (OC)2
b2 = a2 + a2 ⇒ b2 = 2a2
b = √2 a
Most Upvoted Answer
Two chords of length a unit and b unit of a circle make angles 60°...
Let the chord AB = a and chord BC = b makes angle
∠ AOB = 60° and ∠ BOC = 90° at the center ‘O’ of the circle. There, OA = OB = OC = radius
In ΔAOB, ∠ OAB = ∠ OBA
and ∠ AOB = 60°
∴ ∠ OAB + ∠ OBA = 180° – 60° = 120°
⇒ ∠ OAB + ∠ OAB = 120°
⇒ ∠ OAB = 60°
Thus, ∠ OAB = ∠ OBA = ∠ AOB = 60°
∴ AOB is equilateral triangle
Hence, AO = OB = AB = a unit
Now, from ΔBOC, ∠ BOC = 90°, BC = b unit
OB = OC = a unit
(BC)2 = (OB)2 + (OC)2
b2 = a2 + a2 ⇒ b2 = 2a2
b = √2 a
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Two chords of length a unit and b unit of a circle make angles 60° and 90° at the centre of a circle respectively, then the correct relation is (SSC CGL 1st Sit. 2015)a)b = √2 ab)b = 2ac)b = √3ad)Correct answer is option 'A'. Can you explain this answer?
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