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
Answers
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
Here is your answer bro/sis.
