Question -
Answer -
In the given figure,
AB is the diameter, AT is the tangent
and тИаAOQ = 58┬░
To find тИаATQ
Arc AQ subtends тИаAOQ at the centre and тИаABQ at the remaining part of the circle
тИаABQ = 12 тИаAOQ = 12 x 58┬░ = 29┬░
Now in тИЖABT,
тИаBAT = 90┬░ ( OA тКе AT)
тИаABT + тИаATB = 90┬░
=> тИаABT + тИаATQ = 90┬░
=> 29┬░ + тИаATQ = 90┬░
=> тИаATQ = 90┬░- 29┬░ = 61┬░