Advertisements
Advertisements
Question
In the given figure, O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.
Solution
Join OC.
Therefore, PA and PC are the tangents
∴ OA ⊥ PA and OC ⊥ PC
In quadrilateral APCO,
∠APC + ∠AOC = 180°
`=>` 80° + ∠AOC = 180°
`=>` ∠AOC = 100°
∠BOC = 360° – (∠AOB + ∠AOC)
∠BOC = 360° – (140° + 100°)
∠BOC = 360° – 240° = 120°
Now, arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.
∴ `∠BAC = 1/2 ∠BOC`
`∠BAC =1/2 xx 120^circ = 60^circ`
APPEARS IN
RELATED QUESTIONS
From the given figure, prove that : AP + BQ + CR = BP + CQ + AR.
Also show that : AP + BQ + CR = `1/2` × Perimeter of ΔABC.
Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that : ∠PAQ = 2∠OPQ
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate:
i)`∠`QPR .
In the following figure, PQ and PR are tangents to the circle, with centre O. If `∠`QPR = 60°, calculate:
- ∠QOR,
- ∠OQR,
- ∠QSR.
In the given figure, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P. Given ∠SPR = x° and ∠QRP = y°;
Prove that:
- ∠ORS = y°
- write an expression connecting x and y.
PT is a tangent to the circle at T. If ∠ABC = 70° and ∠ACB = 50°; calculate:
- ∠CBT
- ∠BAT
- ∠APT
AB is the diameter and AC is a chord of a circle with centre O such that angle BAC = 30°. The tangent to the circle at C intersects AB produced in D. show that BC = BD.
In the figure; PA is a tangent to the circle, PBC is secant and AD bisects angle BAC. Show that triangle PAD is an isosceles triangle. Also, show that:
`∠CAD = 1/2 (∠PBA - ∠PAB)`
In the given figure, AB is the diameter. The tangent at C meets AB produced at Q. If ∠CAB = 34°, find:
- ∠CBA
- ∠CQB
In the joining figure shown XAY is a tangent. If ∠ BDA = 44°, ∠ BXA = 36°.
Calculate: (i) ∠ BAX, (ii) ∠ DAY, (iii) ∠ DAB, (iv) ∠ BCD.
