Plane Geometry
- In a circle if PQ is the diameter of the circle and R is on the circumference of the circle such that ∠PQR = 30°, then ∠RPQ = ?
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As per the given in question , we draw a figure of circle in which PQ is the diameter
Angle of a semi–circle is a right angle.
Given , ∠PQR = 30°
∴ ∠PRQ = 90°
In ∆ PQR,
∠RPQ + ∠RQP = 90°Correct Option: B
As per the given in question , we draw a figure of circle in which PQ is the diameter
Angle of a semi–circle is a right angle.
Given , ∠PQR = 30°
∴ ∠PRQ = 90°
In ∆ PQR,
∠RPQ + ∠RQP = 90°
⇒ ∠RPQ + 30° = 90°
⇒ ∠RPQ = 90° – 30° = 60°
- If AB = 5 cm, AC = 12 cm and AB ⊥ AC, then the radius of the circumcircle of ∆ ABC is
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According to question , we draw a figure of triangle ABC circumscribes a circle with centre O
Here , AB = 5 cm, AC = 12 cm and AB ⊥ AC
BC = Diameter
From ∆ABC ,
∴ BC = √AB² + AC²
BC = √5² + 12² = √25 + 144
BC = √169 = 13 cmCorrect Option: A
According to question , we draw a figure of triangle ABC circumscribes a circle with centre O
Here , AB = 5 cm, AC = 12 cm and AB ⊥ AC
BC = Diameter
From ∆ABC ,
∴ BC = √AB² + AC²
BC = √5² + 12² = √25 + 144
BC = √169 = 13 cm∴ OB = circum-radius = BC = 13 = 6.5 cm 2 2
- The circumcentre of a triangle ABC is O. If ∠ BAC = 85° and ∠ BCA = 75°, then the value of ∠ OAC is
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According to question , we draw a figure of triangle ABC and O is the circumcentre of a triangle ABC
Given , ∠BAC = 85°
∴ ∠BOC = 2 × 85° = 170°Correct Option: C
According to question , we draw a figure of triangle ABC and O is the circumcentre of a triangle ABC
Given , ∠BAC = 85°
∴ ∠BOC = 2 × 85° = 170°
⇒ ∠OBC = ∠OCB = 5°
∴ ∠OCA = ∠OAC = 75° – 5° = 70°
- ABC is an equilateral triangle and O is its circumcentre, then the ∠AOC is
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As per the given in question , we draw a figure of an equilateral triangle whose circumcentre is O
In ∆ ABC,
∠ BAC = 60°, ∠ ABC = 60°Correct Option: C
As per the given in question , we draw a figure of an equilateral triangle whose circumcentre is O
In ∆ ABC,
∠ BAC = 60°, ∠ ABC = 60°
We can say that the angle subtended by an arc at the centre is twice to that at the circumference by the same arc.
∴ ∠ AOC = 2∠ABC = 120°
- O is the circumcentre of ∆ABC, given ∠BAC = 85° and ∠BCA = 55°, find ∠OAC.
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According to question , we draw a figure of triangle ABC whose the circumcentre is O
Given , ∠BAC = 85°
∴ ∠BOC = 2 × 85° = 170°
∠BCA = 55°
∴ ∠AOB = 2 × 55° = 110°
We know that , ∠AOC + ∠BOC + ∠AOB = 360°
∴ ∠AOC = 360° – 170° – 110° = 360° – 280° = 80°Correct Option: B
According to question , we draw a figure of triangle ABC whose the circumcentre is O
Given , ∠BAC = 85°
∴ ∠BOC = 2 × 85° = 170°
∠BCA = 55°
∴ ∠AOB = 2 × 55° = 110°
We know that , ∠AOC + ∠BOC + ∠AOB = 360°
∴ ∠AOC = 360° – 170° – 110° = 360° – 280° = 80°∴ ∠OAC = ∠OCA = 1 (180° - 80°) 2 ∠OAC = 1 × 100 = 50° 2
