Question

(please teach me how to do this question ,tqsm!!!)​

Answer 1

`\sf\\(i)\ \angle RQS+\angle QRS+\angle QSR=180^o\ \ \ [\textsf{Sum of angles of triangle is 180}^o.]\\or,\ \angle RQS+55^o+90^o=180^o\\or,\ \angle RQS=35^o`

`\sf\\(ii)\ \angle PQR=\angle PQS-\angle RQS=110^o-35^o=75^o\\\\(iii)\ \angle PRQ=\angle PQR=75^o\ \ \ [\textsf{Base angles of isosceles triangle are equal.}]\\\\(iv)\ \angle PRS=\angle PRQ+\angle QRS=75^o+55^o=130^o\\or,\ b=130^o`

`\sf\\(v)\ \angle P+\angle PQR+\angle PRQ=180^o\ \ \ [\textsf{Sum of angles of triangle is 180}^o.]\\or,\ a+75^o+75^o=180^o\\\therefore\ a=30^o`