Plane Geometry
- If PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then∠OAB is equal to
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As per the given in question , we draw a figure of circle with centre O in which PA and PB are tangents
PA = PB (tangents from an exterior point)
OA = OB = radii
Given , ∠APB = 50°
∠APO = ∠OPB = 25°Correct Option: A
As per the given in question , we draw a figure of circle with centre O in which PA and PB are tangents
PA = PB (tangents from an exterior point)
OA = OB = radii
Given , ∠APB = 50°
∠APO = ∠OPB = 25°∠PAB = ∠PBA = 130° = 65° 2
∠OAP = 90°
∴ ∠OAB = 90° – 65° = 25°
- A, B and C are three points on a circle with centre O. The tangent at C meets BA produced to T. If ∠ATC = 30° and∠ACT = 48°, then what is the value of ∠AOB ?
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According to question , we draw a figure circle with centre O in which A, B and C are three points
Here , ∠ATC = 30°;
∠ACT = 48°
From ∆ CAT ,
∴ ∠CAT + ∠ACT + ∠ATC = 180°
∴ ∠CAT = 180° – (30° + 48°) = 180° – 78° = 102°Correct Option: D
According to question , we draw a figure circle with centre O in which A, B and C are three points
Here , ∠ATC = 30°;
∠ACT = 48°
From ∆ CAT ,
∴ ∠CAT + ∠ACT + ∠ATC = 180°
∴ ∠CAT = 180° – (30° + 48°) = 180° – 78° = 102°
∴ ∠OCA = 90° – 48° = 42° = ∠OAC
∴ ∠OAB = 180° – 102° – 42° = 36° = ∠OBA
∴ ∠AOB = 180° – 2 × 36° = 108°
- AB is a diameter of a circle. C is a point on the tangent drawn at A. If AB = 8 cm and AC = 6 cm, then the length of BC is :
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On the basis of question we draw a figure of a circle with centre O ,
Given that , AB = 8 cm and AC = 6 cm
AB ⊥ AC
∴ ∠BAC = 90°
From ∆ BAC ,
∴ BC = √AB² + AC²Correct Option: A
On the basis of question we draw a figure of a circle with centre O ,
Given that , AB = 8 cm and AC = 6 cm
AB ⊥ AC
∴ ∠BAC = 90°
From ∆ BAC ,
∴ BC = √AB² + AC²
BC = √8² + 6²
BC = √64 + 36 = √100 = 10 cm.
- From an external point two tangents to a circle are drawn. The chord passing through the points of contact subtends an angle 72° at the centre. The angle between the tangents is
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As per the given in question , we draw a figure of circle in which two tangents are drawn from an external point
From figure , OA = OB = radii of circle
Given , ∠AOB = 72°;
∠OBP = ∠OAP = 90°
In ∠OAB,
∠OAB = ∠OBA
∴ 2 ∠OAB = 180° – 72° = 108°Correct Option: C
As per the given in question , we draw a figure of circle in which two tangents are drawn from an external point
From figure , OA = OB = radii of circle
Given , ∠AOB = 72°;
∠OBP = ∠OAP = 90°
In ∠OAB,
∠OAB = ∠OBA
∴ 2 ∠OAB = 180° – 72° = 108°⇒ ∠OAB = 108° = 54° 2
∴ ∠PBA = ∠PAB = 90° – 54° = 36°
∴ ∠BPA = 180° – 2 × 36°
∠BPA = 180° – 72° = 108°
- PQ is a tangent to the circle at T. If TR = TS where R and S are points on the circle and ∠RST = 65°, the ∠PTS = ?
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On the basis of question we draw a figure of a circle with centre O and PQ is a tangent
Here , TR = TS
∠SRT = ∠TSR = 65°
∴ ∠STR = 180° – 130° = 50°Correct Option: C
On the basis of question we draw a figure of a circle with centre O and PQ is a tangent
Here , TR = TS
∠SRT = ∠TSR = 65°
∴ ∠STR = 180° – 130° = 50°∴ ∠PTR = ∠STQ = 130° = 65° 2
∴ ∠PTS = 65° + 50° = 115°
