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The Given Figure Shows a Semi-circle with Centre O and Diameter Pq. If Pa = Ab and ∠Bcq = 140°; Find Measures of Angles Pab and Aqb. Also, Show that Ao is Parallel to Bq. - Mathematics

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Question

The given figure shows a semi-circle with centre O and diameter PQ. If PA = AB and ∠BCQ =140°; Find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.

Solution

Join PB.
i) In cyclic quadrilateral PBCQ,
`∠`BPQ + `∠`BCQ = 180°

⇒  `∠`BPQ + 140°  =  180°
⇒ `∠`BPQ = 40°  ……. (1)
Now in ΔPBQ,
`∠`PBQ  + BPQ + `∠`BQP = 180°
⇒ 90° + 40° + `∠`BQP = 180°
⇒  `∠`BQP = 50°
In cyclic quadrilateral PQBA,
`∠`PQB + `∠`PAB = 180°
⇒ 50° + `∠`PAB =  180°
⇒ `∠`PAB = 130°

 

ii) Now in  ΔPAB,
`∠`PAB + `∠`APB  + ABP = 180°
⇒ 130° + `∠`APB  + `∠`ABP = 180°
⇒  `∠`APB + `∠`ABP = 50°
But
`∠`APB = `∠`ABP ( ∵ PA = PB)
∴  `∠`APB = `∠`ABP = 25°
`∠`BAQ = `∠`BPQ = 40°
`∠`APB = 25°  = `∠`AQB  (angles in the same segment)
∴  `∠`AQB = 25° ………. (2)

 

iii) Arc AQ subtends  `∠`AOQat the centre and `∠`APQ at the remaining part of the circle.
We have,
`∠`APQ =  `∠`APB + `∠`BPQ ……. (3)
From (1), (2) and (3), we have
`∠`APQ = 25° + 40° = 65°
∴ `∠`AOQ =  2`∠`APQ  = 2 × 65° = 130°
Now in ΔAOQ,
`∠`OAQ =  `∠`OQA = ( ∵ OA = OQ)
But
`∠` OAQ  + `∠`OQA + `∠`AOQ  = 180°
 ⇒ `∠`OAQ  + `∠`OAQ  +130° = 180°
⇒ 2 `∠`OAQ  = 50°
⇒  `∠`OAQ = 25°
 ∴ `∠`OAQ  =  `∠`AQB
But these are alternate angles.
Hence, AO is parallel to BQ.

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APPEARS IN

 Selina Solution for Concise Mathematics for Class 10 ICSE (2020 (Latest))
Chapter 18: Tangents and Intersecting Chords
Exercise 18 (C) | Q: 23 | Page no. 286
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The Given Figure Shows a Semi-circle with Centre O and Diameter Pq. If Pa = Ab and ∠Bcq = 140°; Find Measures of Angles Pab and Aqb. Also, Show that Ao is Parallel to Bq. Concept: Cyclic Properties.
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