3. AB is a diameter of the circle with centre O.
BT is a tangent. AC produced meets tangent
BT at T. If ABC = 38°, find T.
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Answer:
∠ACT=48
o
,∠ATC=36
o
∴∠CAT=180
o
−(48
o
+36
o
)
=180
o
−84
o
=96
o
∴CAB=180
o
−96
o
=84
o
∴CAB=180
o
−96
o
=84
o
Now,∠ABC=∠ACT=48
o
∴∠BCA=180
o
−(∠ABC+∠CAB)∴∠BCA=180
o
−(∠ABC+∠CAB)
=180
o
−(48
o
+84
o
)=48
o
∴∠BOA=2×∠BCA=2×48
o
=96
o
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