In figure O is the centre of the circle and TP is the tangent to the circle from an external point P .Prove this above question
Answers
Step-by-step explanation:
AB is the chord passing through the center,
So, AB is the diameter
ince angle in a semi-circle is a right angle
∠APB= 90°
By using alternate segment theorem
We have ∠APB = ∠PAT = 30°
Now, in ⧍APB
∠BAP + ∠APB + ∠BAP = 180′ (Angle sum property of triangle)
∠BAP = 180° – 90° – 30° = 60°
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
60° = 30° + ∠PTA
∠PTA = 60° – 30° = 30°
We know that sides opposite to equal angles are equal
AP = AT
In right triangle ABP:
=> sin ABP=AP/BA
=> sin30° = AT/BA {since AP = AT}
=>1/2 = AT/BA
=> BA/AT = 2/1
=> BA : AT = 2 : 1:
AB is the chord passing through the center,
So, AB is the diameter
Since angle in a semi-circle is a right angle
APB= 90°
By using alternate segment theorem
We have ∠APB = ∠PAT = 30°
Now, in ⧍APB
∠BAP + ∠APB + ∠BAP = 180′ (Angle sum property of triangle)
∠BAP = 180° – 90° – 30° = 60°
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
60° = 30° + ∠PTA
∠PTA = 60° – 30° = 30°
We know that sides opposite to equal angles are equal
AP = AT
In right triangle ABP:
=> sin ABP=AP/BA
=> sin30° = AT/BA {since AP = AT}
=>1/2 = AT/BA
=> BA/AT = 2/1
=> BA : AT = 2 : 1
AB is the chord passing through the center,
So, AB is the diameter
Since angle in a semi-circle is a right angle
∠APB= 90°
By using alternate segment theorem
We have ∠APB = ∠PAT = 30°
Now, in ⧍APB
∠BAP + ∠APB + ∠BAP = 180′ (Angle sum property of triangle)
∠BAP = 180° – 90° – 30° = 60°
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
60° = 30° + ∠PTA
∠PTA = 60° – 30° = 30°
We know that sides opposite to equal angles are equal
AP = AT
In right triangle ABP:
=> sin ABP=AP/BA
=> sin30° = AT/BA {since AP = AT}
=>1/2 = AT/BA
=> BA/AT = 2/1
=> BA : AT = 2 : 1:
AB is the chord passing through the center,
So, AB is the diameter
Since angle in a semi-circle is a right angle
∠APB= 90°
By using alternate segment theorem
We have ∠APB = ∠PAT = 30°
Now, in ⧍APB
∠BAP + ∠APB + ∠BAP = 180′ (Angle sum property of triangle)
∠BAP = 180° – 90° – 30° = 60°
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
60° = 30° + ∠PTA
∠PTA = 60° – 30° = 30°
We know that sides opposite to equal angles are equal
AP = AT
In right triangle ABP:
=> sin ABP=AP/BA
=> sin30° = AT/BA {since AP = AT}
=>1/2 = AT/BA
=> BA/AT = 2/1
=> BA : AT = 2 : 1
Hence proved
Hence proved
The answer of ur question is here
Angle AOP = 30*2 =60
angle OAP =180-30-90 = 60
therefore OP = PA
also angle ATP = angle APT = 30
therefore AP=AT=OP=OA
hence BA = 2 OA =2 AT
BA:AT=2:1
So , It is proved
