Advertisements
Advertisements
प्रश्न
In the given figure, AC is a tangent to circle at point B. ∆EFD is an equilateral triangle and ∠CBD = 40°. Find:
- ∠BFD
- ∠FBD
- ∠ABF
उत्तर
Given - In Diagram, ∆FED is an equilateral triangle, ∠CBD = 40°
To Find -
- ∠BFD,
- ∠FBD,
- ∠ABF
(a) ∠BFD = ∠CBD = 40° ....[Alternate segment theorem]
(b) ∠FBD is opposite to ∠FED.
So, It is a cyclic Quadrilateral. So, ∠FBD and ∠FED will be Supplementary to each other.
∴ ∠FBD + ∠FED = 180° ....[Cyclic Quadrilateral]
∠FBD + 60° = 180°
∠FBD = 180° - 60°
∠FBD = 120°
(c) AC is a line segment.
So, ∠ABF = 180° - (120° + 40°) ....[Line Segment]
∠ABF = 180° - 100°
∠ABF = 20°
APPEARS IN
संबंधित प्रश्न
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate :
- ∠QOR
- ∠QPR;
given that ∠A = 60°.
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 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.
Tangent at P to the circumcircle of triangle PQR is drawn. If the tangent is parallel to side, QR show that ΔPQR is isosceles.
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 following figure, PQ and PR are tangents to the circle, with centre O. If ∠ QPR = 60° , calculate:
∠ QSR
In Fig. AP is a tangent to the circle at P, ABC is secant and PD is the bisector of ∠BPC. Prove that ∠BPD = `1/2` (∠ABP - ∠APB).
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.
The figure shows a circle of radius 9 cm with 0 as the centre. The diameter AB produced meets the tangent PQ at P. If PA = 24 cm, find the length of tangent PQ:
