Mathematics
In the figure (ii) given below, AP and BP are tangents to the circle with centre O. Given ∠APB = 60°, calculate :
(i) ∠AOB
(ii) ∠OAB
(iii) ∠ACB
Circles
16 Likes
Answer
(i) From figure,
OA ⊥ AP and OB ⊥ BP (∵ OA and OB are the radii and AP and BP are tangents.)
Now in quadrilateral AOBP,
∠P = 60°, ∠OAP = 90° and ∠OBP = 90°.
∠P + ∠OAP + ∠OBP + ∠AOB = 360°
60° + 90° + 90° + ∠AOB = 360°
240° + ∠AOB = 360°
∠AOB = 360° - 240°
∠AOB = 120°.
Hence, the value of ∠AOB = 120°.
(ii) Join AB as shown in the figure below:
Considering △OAB,
The triangle is isosceles as OA = OB = radii of the circle so, ∠OAB = ∠OBA = x.
Since sum of angles in a triangle = 180.
In △OAB,
⇒ ∠AOB + ∠OAB + ∠OBA = 180°
⇒ 120° + x + x = 180°
⇒ 120° + 2x = 180°
⇒ 2x = 180° - 120°
⇒ 2x = 60°
⇒ x = 30°
Hence, the value of ∠OAB = 30°.
(iii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.
∴ ∠AOB = 2∠ACB (∵ angle subtended at centre by an arc is double the angle subtended at remaining part of circle)
120° = 2∠ACB
∠ACB = = 60°.
Hence, the value of ∠ACB = 60°.
Answered By
10 Likes
Related Questions
In the figure (i) given below, O is the centre of the circumcircle of triangle XYZ. Tangents at X and Y intersect at T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY.
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at P, Q and R respectively. Calculate :
(i) ∠QOR
(ii) ∠QPR, given that ∠A = 60°.
In the figure (ii) given below, O is the center of the circle and PT is the tangent to the circle at P. Given ∠QPT = 30°, calculate
(i) ∠PRQ
(ii) ∠POQ.
In the figure (i) given below, AB is a diameter. The tangent at C meets AB produced at Q, ∠CAB = 34°. Find :
(i) ∠CBA
(ii) ∠CQA