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In the given figure TBP TCQ are tangents to the circle whose centre is O.? PBA = 60 and ? ACQ = 70. Find angle BAC and angle BTC.

see .. u have to join OB , OA AND OC ... ang.ABO=ang.BAO =90-60 =30 (these angle r equal bcz radius are equal)

now ang. OCA = ANG. OAC =90-70= 20

ANG. BAC=ANG.BAO +ANGL.CAO= 50

2 ANG. BAC= ANG. BOC = 2x50 =100

ANG.BTC = 180-100 = 80

hope u got it

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figure????
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Kanika, why ANG.BTC=180-100 ?
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join OA , OB AND OC?in triangle OABangle 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?? ?
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Where is figure?
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ACB = ABP = 60° ABC = ACQ = 70°  BAC = 180° – (60° + 70°) = 50°  BOC = 100°  BTC = 180° – 100° = 80°
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