Question

in the given figure,two chords PQ and RS of the circle intersect at point T,such that angle STQ =123 degree and angle TQR =17 degree. find angle SPQ​

Answer 1

$\bold{Answer}$

Given: m(arc PCR) = 26°, m(arc QDS) = 48°

By Inscribed Angle Theorem,we get

(i) ∠PQR = 1/2 m(arc PCR)

∠PQR = 1/2 x 26º

∠PQR = ∠AQR = 13º ...(1)

(ii) ∠SPQ = 1/2m(arc QDS)

∠SPQ = 1/2 x 48º

∠SPQ = 24º ...(2)

(iii) In ΔAQR, by the Remote Interior Angle theorem,

∠RAQ + ∠AQR = ∠SRQ

∠SRQ = ∠SPQ ...(Angles subtended by the same arc)

i.e. ∠RAQ + ∠AQR = ∠SPQ

m ∠RAQ = m ∠SPQ - m∠AQR

m ∠RAQ = 24º - 13º ...[From (1) and (2)]

m ∠RAQ = 11º

Answer 2

Answer:

$\huge \underline \pink {answer}$

Given: m(arc PCR) = 26°, m(arc QDS) = 48°

By Inscribed Angle Theorem,we get

(i) ∠PQR = 1/2 m(arc PCR)

∠PQR = 1/2 x 26º

∠PQR = ∠AQR = 13º ...(1)

(ii) ∠SPQ = 1/2m(arc QDS)

∠SPQ = 1/2 x 48º

∠SPQ = 24º ...(2)

(iii) In ΔAQR, by the Remote Interior Angle theorem,

∠RAQ + ∠AQR = ∠SRQ

∠SRQ = ∠SPQ ...(Angles subtended by the same arc)

i.e. ∠RAQ + ∠AQR = ∠SPQ

m ∠RAQ = m ∠SPQ - m∠AQR

m ∠RAQ = 24º - 13º ...[From (1) and (2)]

m ∠RAQ = 11º