Question

In the given figure PQ is a diameter of a circle with
centre O and PT is a tangent at P. QT meets the
circle at R. If ZPOR = 72°, find ZPTR.
​

Answer 1

(i) ∠QPR  ∴ PQ is a diameter  ∴ ∠PRQ = 900  [Angle in a semi-circle is 900]  In ΔPQR,  ∠QPR + ∠PRQ + ∠PQR = 1800 [Angle Sum Property of a triangle]  => ∠QPR + 900 + 650= 1800 => ∠QPR + 1550 = 1800 => ∠QPR = 1800 - 1550  => ∠QPR = 250.  (ii) ∠PRS  ∴ PQRS is a cyclic quadrilateral  ∴ ∠PSR + ∠PQR = 1800 [∴ Opposite angles of a cyclic quadrilateral are supplementary]  => ∠PSR + 650 = 1800  => ∠PSR = 1800- 650  => ∠PSR = 1150  In DPSR,  ∠PSR + ∠SPR + ∠PRS = 1800 [Angles Sum Property of a triangle]  => 1150 + 400 + ∠PRS = 1800  => 1150 + ∠PRS = 1800  => ∠PRS = 1800 - 1550  => ∠PRS = 250  (iii) ∠QPM  ∴ PQ is a diameter  ∴ ∠PMQ = 900 [∴ Angle in a semi - circle is 900]  In ΔPMQ,  ∠PMQ + ∠PQM + ∠QPM = 1800 [Angle sum Property of a triangle]  => 900 + 500 + ∠QPM = 1800  => 1400 + ∠QPM = 1800 => ∠QPM = 1800 - 1400  => ∠QPM = 40

Answer 2

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(i) ∠QPR  ∴ PQ is a diameter  ∴ ∠PRQ = 900  [Angle in a semi-circle is 900]  In ΔPQR,  ∠QPR + ∠PRQ + ∠PQR = 1800 [Angle Sum Property of a triangle]  => ∠QPR + 900 + 650= 1800 => ∠QPR + 1550 = 1800 => ∠QPR = 1800 - 1550  => ∠QPR = 250.  (ii) ∠PRS  ∴ PQRS is a cyclic quadrilateral  ∴ ∠PSR + ∠PQR = 1800 [∴ Opposite angles of a cyclic quadrilateral are supplementary]  => ∠PSR + 650 = 1800  => ∠PSR = 1800- 650  => ∠PSR = 1150  In DPSR,  ∠PSR + ∠SPR + ∠PRS = 1800 [Angles Sum Property of a triangle]  => 1150 + 400 + ∠PRS = 1800  => 1150 + ∠PRS = 1800  => ∠PRS = 1800 - 1550  => ∠PRS = 250  (iii) ∠QPM  ∴ PQ is a diameter  ∴ ∠PMQ = 900 [∴ Angle in a semi - circle is 900]  In ΔPMQ,  ∠PMQ + ∠PQM + ∠QPM = 1800 [Angle sum Property of a triangle]  => 900 + 500 + ∠QPM = 1800  => 1400 + ∠QPM = 1800 => ∠QPM = 1800 - 1400  => ∠QPM = 40

Thanks!