Question

HELP ASAP I NEED THIS PROBLEM FAST

Answer 1

`\sf\\\textsf{Here, AB and AC are equal. Hence, }\angle B\textsf{ and }\angle C\textsf{ are the base angles of isosceles}\\\textsf{triangle ABC. So they are equal.}\\\therefore \angle B=\angle C\\\textsf{or, }(3y-10)^o=(y+30)^o\\\textsf{or, }3y-y=30^o+10^o\\\textsf{or, }2y=40^o\\\textsf{or, }y=20^o\\\\Now,\\(i )\ \angle B=\angle C=3y-10^o=3(20^o)-10^o=50^o`

`\sf\\\textsf{Also,}\\(ii)\ \angle A+\angle B+\angle C=180^o\ \ \ [\textsf{Sum of angles of triangle is 180}^o.]\\\textsf{or, }\angle A+50^o+50^o=180^o\\\textsf{or, }\angle A+100^o=180^o\\\textsf{or, }\angle A=80^o`

`\textsf{AB and AC are already given equal, }\\\textsf{So AB = AC}\\\sfor,\ 2x-1=x+5\\\textsf{or, }x=6\\Finally,\\(iii)\ AB=AC=x+5=6+5=11`