Question

If QR RT
find the value of x.
R: (2x - 10)⁰
S: 110°
T: 64°

Answer 1

`\sf\\Solution:\\\\(i)\ \angle QSR+\angle TSR=180^o\ \ \ [\textsf{Sum of angles in straight line is 180}^o.]\\or,\ 110^o+\angle TSR=180^o\\or,\ \angle TSR=70^o`

`\sf\\(ii)\ \angle TSR+\angle STR+\angle SRT=180^o\ \ \ [\textsf{Sum of angles of triangle is 180}^o]\\or,\ 70^o+64^o+\angle SRT=180^o\\or,\ \angle SRT=46^o`

`\sf\\(iii)\ \angle QRS+\angle SRT=\angle QRT\ \ \ [\textsf{Whole part axiom}]\\or,\ (2x-10)^o+46^o=90^o\ \ \ [QR\perp RT, therefore\ \angle QRT=90^o]\\or,\ 2x+36^o=90^o\\or,\ 2x=54^o\\or,\ x=27^o`