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

[chijioke]

This is my effort in solving the problem.


Here is my work:

[math]S\widehat{Q}R=44\degree\ ~(\text{>s in alternate segment})[/math]
[math]P\widehat{S}R=P\widehat{S}Q + Q\widehat{S}R[/math]
[math]S\widehat{Q}R= Q\widehat{S}R=44\degree ~(\text{base >s of an isosceles triangle})[/math]
[math]Q\widehat{R}S=180-(2\times 44) = 92\degree~ (\text{sum of >s in a triangle })[/math]
[math]Q\widehat{R}S= Q\widehat {P}S=92\degree ~(\text{opposite >s of a cyclic quadrilateral})[/math]
[math]P\widehat{S}Q=180-(92+31) = 57\degree[/math]
Since [math]P\widehat{S}R=P\widehat{S}Q + Q\widehat{S}R[/math]
[math]P\widehat{S}R =57\degree +44\degree =101\degree[/math]
Is my solution correct?

 

[Dr.Peterson]

[math]Q\widehat{R}S= Q\widehat {P}S=92\degree ~(\text{opposite >s of a cyclic quadrilateral})[/math]

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.

 

[chijioke]

Opposite angles of a cyclic quadrilateral are not congruent!

Mistake! They are supplementary instead.

[math]S\widehat{Q}R=44\degree\ ~(\text{>s in alternate segment})[/math]
[math]P\widehat{S}R=P\widehat{S}Q + Q\widehat{S}R[/math]
[math]S\widehat{Q}R= Q\widehat{S}R=44\degree ~(\text{base >s of an isosceles triangle})[/math]
[math]Q\widehat{R}S=180-(2\times 44) = 92\degree~ (\text{sum of >s in a triangle })[/math]
[math]Q\widehat{P}S= 180-92=88\degree ~(\text{opposite >s of a cyclic quadrilateral})[/math]
[math]P\widehat{S}Q=180-(88+31) = 61\degree~(\text{sum of >s in a triangle})[/math]
Since [math]P\widehat{S}R=P\widehat{S}Q + Q\widehat{S}R[/math]
[math]P\widehat{S}R =61\degree +44\degree =105\degree[/math]I think it is now correct.

 

[The Highlander]

[math]P\widehat{S}R =61\degree +44\degree =105\degree[/math]I think it is now correct.

Or, “a picture (or sketch?) is worth a thousand words” ?



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

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

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