Question

In fig. 2, AOB is the diameter of the circle with center O. The tangent at a point T on the circle, meets AB produced at P. If /_BAT =30°, find /_ TPA​

Answer 1

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Answer 2

Answer:

30° is the required angle of ∠TPA .

Step-by-step explanation:

Explanation:is

Given, AOB is the diameter of the circle

TP is tangent and ∠ BAT = 30°

Step 1:

As we know if two sides of triangle are equal then the angle opposite to the same two sides are also equal .

Therefore in ΔATO  

OA =OT (radius of the circle )

∴ ∠OAT =∠OTA = 30 °

Now in ΔATO

∠OAT +∠OTA+∠TOA = 180°       (sum of angle of a triangle are 180°)

⇒30 +30 +∠TOA = 180

⇒ ∠TOA = 180 -60 =120°

Step2:

∠TOA+∠TOB = 180°    (straight line )

⇒∠TOB = 180-120       (120 is the angle of ∠TOA)

⇒∠TOB = 60 °

Now in triangle ΔBTO

OB = TO       (radius of circle )

∴∠OTB=∠OBT  =  $\alpha$   (let $\alpha$ be the angle of ∠OTBand ∠OBT)

In ΔBTO

∠OTB+∠OBT+∠TOB = 180 ° (sum of angle of triangle are 180 )

⇒$\alpha +\alpha +60 =180$

⇒$2\alpha =180-60=120$

⇒$\alpha = 60$ = ∠OTB=∠OBT

Step 3:

Given that ,tangent at a point T

So line OT perpendicular on T

∴∠OTP = 90°

In ΔOTP we have ∠TOP = 60° and ∠OTP = 90°

Now ,∠TOP+∠OPT+∠PTO = 180 °

⇒ 60 +∠OPT +90 = 180

⇒∠OPT = 180-150=30°

∴∠OPT=∠TPA = 30

Final answer :

Hence , the angle of ∠TPA is 30 °