Calculate angle PSR: In the drawing, ST is a tangent to a circle PQRS....

chijioke

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IMG_20230604_133205.jpg
This is my effort in solving the problem.
IMG_20230604_132527.jpg

Here is my work:

SQ^R=44°  (>s in alternate segment)S\widehat{Q}R=44\degree\ ~(\text{>s in alternate segment})
PS^R=PS^Q+QS^RP\widehat{S}R=P\widehat{S}Q + Q\widehat{S}R
SQ^R=QS^R=44° (base >s of an isosceles triangle)S\widehat{Q}R= Q\widehat{S}R=44\degree ~(\text{base >s of an isosceles triangle})
QR^S=180(2×44)=92° (sum of >s in a triangle )Q\widehat{R}S=180-(2\times 44) = 92\degree~ (\text{sum of >s in a triangle })
QR^S=QP^S=92° (opposite >s of a cyclic quadrilateral)Q\widehat{R}S= Q\widehat {P}S=92\degree ~(\text{opposite >s of a cyclic quadrilateral})
PS^Q=180(92+31)=57°P\widehat{S}Q=180-(92+31) = 57\degree
Since PS^R=PS^Q+QS^RP\widehat{S}R=P\widehat{S}Q + Q\widehat{S}R
PS^R=57°+44°=101°P\widehat{S}R =57\degree +44\degree =101\degree
Is my solution correct?
 
QR^S=QP^S=92° (opposite >s of a cyclic quadrilateral)Q\widehat{R}S= Q\widehat {P}S=92\degree ~(\text{opposite >s of a cyclic quadrilateral})

Opposite angles of a cyclic quadrilateral are not congruent!

But the rest of the work is well done. I approached it a little differently, but your approach will work with that one fix.
 
Opposite angles of a cyclic quadrilateral are not congruent!
Mistake! They are supplementary instead.

SQ^R=44°  (>s in alternate segment)S\widehat{Q}R=44\degree\ ~(\text{>s in alternate segment})
PS^R=PS^Q+QS^RP\widehat{S}R=P\widehat{S}Q + Q\widehat{S}R
SQ^R=QS^R=44° (base >s of an isosceles triangle)S\widehat{Q}R= Q\widehat{S}R=44\degree ~(\text{base >s of an isosceles triangle})
QR^S=180(2×44)=92° (sum of >s in a triangle )Q\widehat{R}S=180-(2\times 44) = 92\degree~ (\text{sum of >s in a triangle })
QP^S=18092=88° (opposite >s of a cyclic quadrilateral)Q\widehat{P}S= 180-92=88\degree ~(\text{opposite >s of a cyclic quadrilateral})
PS^Q=180(88+31)=61° (sum of >s in a triangle)P\widehat{S}Q=180-(88+31) = 61\degree~(\text{sum of >s in a triangle})
Since PS^R=PS^Q+QS^RP\widehat{S}R=P\widehat{S}Q + Q\widehat{S}R
PS^R=61°+44°=105°P\widehat{S}R =61\degree +44\degree =105\degreeI think it is now correct.
 
Last edited:
PS^R=61°+44°=105°P\widehat{S}R =61\degree +44\degree =105\degreeI think it is now correct.✔️
Or, “a picture (or sketch?) is worth a thousand words” ?

Cyclic Quadrilateral.jpg

      SQ^R=44°  \displaystyle S\widehat{Q}R=44\degree\ ~ (Angles in alternate segment.)

    PQ^R=75° \displaystyle \implies P\widehat{Q}R = 75°~ (31° + 44°)

    PS^R=105° \displaystyle \implies P\widehat{S}R = \underline{\underline{105°}}~ (180° - 75°; Supplementary Angles.)
 
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