Question

o is the centre of the circle and TP is the tangent to the centre from an external point T. IF angle PBT= 30 degree prove that BA; AT = 2; 1

Answer 1

Answer:

BA : AT = 2r : r = 2 : 1 .

Step-by-step explanation:

First let us join the center of the circle, O to the point P.

We know that ∠ABP = 30°.

Since AB is the diameter, ∠APB = 90°.

From the rule that sum of all angles in a triangle = 180°

∠ABP + ∠APB + ∠PAB = 180 °

⇒ 30° + 90° + ∠PAB = 180 °

⇒∠PAB = 60°

We can see that OP and OA are both the radius of the circle. Therefore

OP = OA = radius, r

Therefore ΔOPA is an isosceles triangle with equal sides OP, OA .

∠PAB = ∠OPA = 60°

From the rule sum of all angles in a triangle = 180°

∠OPA + ∠OAP + ∠AOP = 180°

⇒     60° + ∠PAB + ∠AOP = 180

⇒     60 + 60 + ∠AOP = 180

⇒     ∠AOP = 60°

ΔOPA is an equilateral triangle as it can be seen that all its angles are 60°.

From the above we can say that OP =  OA  = PA

Now, ∠OPT will also be equal to 90°, because PT is a tangent to the circle and OP is the radius of the circle.

∠OPA + ∠APT = 90

⇒ 60 + ∠APT = 90

⇒ ∠APT = 30°

We can see that  

∠PAB + ∠PAT = 180°

⇒ 60° + ∠PAT = 180°

⇒ ∠PAT = 120°

Now, looking at △APT , we can see that from the rule of triangles that the sum of all angles in a triangle is equal to 180°

∠APT + ∠PAT + ∠PTA = 180°

30° + 120° + ∠PTA = 180°

∠PTA = 30°

Looking again at △APT, we see that

∠APT = ∠PTA = 30°

AT = PA  = OA = OP = radius, r  ,    because △OPA is equilateral

AT =  radius, r

Now,

BA = OA + OB = r + r = 2r,   where r  is the radius.

BA : AT = 2r : r = 2 : 1

Answer 2

Here is your answer bro/sis.