TBP and TCQ tangents to a circle whose Centre is O angle PBA is equal to 60 degree and Angle ACQ is equal to 70 degree determine angle BAC and angle BTC
Answers
Answer:
join OA , OB AND OC in triangle OAB angle OBA = 90 - 60 = 30 OB = OA (radii) angle OAB = 30 angle AOB = 180 - ( 30 + 30) = 120 in triangle OAC angle OCA = 90 - 70 = 20 OC = OA (RADII) angle OCA = OAC = 20 angle AOC = 180 - (20 +20) 140 thus angle BAC = 50 angle BTC = 180 - 50 = 130
Answer:
Step-by-step explanation:
Mistake in the question. we have to find ∠AQB.
FIGURE IS IN THE ATTACHMENT.
SOLUTION:
Given :
AB || PR & LQ ⟂PR [ OQ ⟂PR = LQ ⟂PR]
LQ ⟂AB = OL ⟂ AB= OL bisect chord AB(AL = BL)
In ∆ ALQ & ∆ BLQ
AL = BL [ proved above]
∠ALQ = ∠BLQ [each angle is 90°]
LQ = LQ [ Common]
∆ ALQ ≅∆ BLQ [SAS Criterion]
∠LQA = ∠LQB [ By CPCT]
∠LQB = 90° - 70° = 20°
∠LQA = ∠LQB= 20°
∠AQB=∠LQA +∠LQB
∠AQB= 20° + 20 ° = 40 °
Hence , ∠AQB is 40°.
HOPE THIS WILL HELP YOU...