The given figure shows a semi-circle with centre O and diameter PQ. If PA = AB and ∠ BCQ = 140o; find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.
The given figure shows a semi-circle with centre O and diameter PQ. If PA = AB and ∠ BCQ = 140o; find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.
Please find below the solution to the asked query:
We have our diagram , As :
Here ∠ BCQ = 140° ( given )
And QCBP is a cyclic quadrilateral ( As all vertices lies on circumference of circle ) . We know opposite angles are supplementary , So
∠ BPQ + ∠ BCQ = 180° , Substitute given value we get
∠ BPQ + 140° = 180° ,
∠ BPQ = 40° --- ( 1 )
And
∠ PBQ = 90° , that is angle in semicircle .
Now from angle sum property in triangle we get in triangle PBQ
∠ BPQ + ∠ PBQ + ∠ BQP = 180° , Substitute above value and from equation 1 we get
40° + 90° + ∠ BQP = 180° ,
∠ BQP = 50°
And ABQP is a cyclic quadrilateral ( As all vertices lies on circumference of circle ) . We know opposite angles are supplementary , So
∠ PAB + ∠ BQP = 180° , Substitute value from above equation , we get
∠ PAB + 50° = 180°
∠ PAB = 130° ( Ans )
Given PA = AB so from base angle theorem in triangle PAB we get
∠ APB = ∠ ABP --- ( 2 )
And from angle sum property of triangle we get in triangle PAB
∠ APB + ∠ ABP + ∠ PAB = 180° , Substitute above value and from equation 2 and and from above answer , we get
∠ APB + ∠ APB + 130° = 180° ,
2 ∠ APB = 50° ,
∠ APB = 25° , So
∠ APB = ∠ ABP = 25° --- ( 3 )
And
∠ APQ = ∠ APB + ∠ BPQ , Now we substitute values from equation 1 and 3 and get
∠ APQ = 25° + 40°
∠ APQ = 65°
And from angle sum property of triangle we get in triangle PAQ
∠ APQ + ∠ PAQ + ∠ AQP = 180° , So
65° + 90° + ∠ AQP = 180° , As we show ∠APQ = 65° and ∠ PAQ = 90° as that is angle in semicircle .
∠ AQP = 25° ,
∠ AQB = ∠ BQP - ∠ AQP , Substitute values as we show in our above solution and get
∠ AQB = 50° - 25 °
∠ AQB = 25° ( Ans )
And OA = OP ( Radius of given semicircle ) so from base angle theorem in triangle AOP we get
∠ PAO = ∠ APO , So
∠ PAO = 65° , As ∠ APO = ∠ APQ , same angle and we show ∠ APQ = 65°
And
∠ OAQ = ∠ PAQ - ∠ PAO , Substitute values as we show in our above solution and get
∠ OAQ = 90° - 65°
∠ OAQ = 25° , Then
∠ OAQ = ∠ ABQ --- ( 4 ) ( As we show ∠ ABQ = 25° )
We know ∠ OAQ and ∠ ABQ are alternate interior angles so equation 4 only be true if we have :
AO | | BQ ( Hence proved )
Hope
Please find below the solution to the asked query:
We have our diagram , As :
Here ∠ BCQ = 140° ( given )
And QCBP is a cyclic quadrilateral ( As all vertices lies on circumference of circle ) . We know opposite angles are supplementary , So
∠ BPQ + ∠ BCQ = 180° , Substitute given value we get
∠ BPQ + 140° = 180° ,
∠ BPQ = 40° --- ( 1 )
And
∠ PBQ = 90° , that is angle in semicircle .
Now from angle sum property in triangle we get in triangle PBQ
∠ BPQ + ∠ PBQ + ∠ BQP = 180° , Substitute above value and from equation 1 we get
40° + 90° + ∠ BQP = 180° ,
∠ BQP = 50°
And ABQP is a cyclic quadrilateral ( As all vertices lies on circumference of circle ) . We know opposite angles are supplementary , So
∠ PAB + ∠ BQP = 180° , Substitute value from above equation , we get
∠ PAB + 50° = 180°
∠ PAB = 130° ( Ans )
Given PA = AB so from base angle theorem in triangle PAB we get
∠ APB = ∠ ABP --- ( 2 )
And from angle sum property of triangle we get in triangle PAB
∠ APB + ∠ ABP + ∠ PAB = 180° , Substitute above value and from equation 2 and and from above answer , we get
∠ APB + ∠ APB + 130° = 180° ,
2 ∠ APB = 50° ,
∠ APB = 25° , So
∠ APB = ∠ ABP = 25° --- ( 3 )
And
∠ APQ = ∠ APB + ∠ BPQ , Now we substitute values from equation 1 and 3 and get
∠ APQ = 25° + 40°
∠ APQ = 65°
And from angle sum property of triangle we get in triangle PAQ
∠ APQ + ∠ PAQ + ∠ AQP = 180° , So
65° + 90° + ∠ AQP = 180° , As we show ∠APQ = 65° and ∠ PAQ = 90° as that is angle in semicircle .
∠ AQP = 25° ,
∠ AQB = ∠ BQP - ∠ AQP , Substitute values as we show in our above solution and get
∠ AQB = 50° - 25 °
∠ AQB = 25° ( Ans )
And OA = OP ( Radius of given semicircle ) so from base angle theorem in triangle AOP we get
∠ PAO = ∠ APO , So
∠ PAO = 65° , As ∠ APO = ∠ APQ , same angle and we show ∠ APQ = 65°
And
∠ OAQ = ∠ PAQ - ∠ PAO , Substitute values as we show in our above solution and get
∠ OAQ = 90° - 65°
∠ OAQ = 25° , Then
∠ OAQ = ∠ ABQ --- ( 4 ) ( As we show ∠ ABQ = 25° )
We know ∠ OAQ and ∠ ABQ are alternate interior angles so equation 4 only be true if we have :
AO | | BQ ( Hence proved )
Hope
