Question

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​

Answer 1

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 2

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...

sorry i change angle name