Question

Please don't give useless answer
If any one give useless answer, I will report her answer​

Answer 1

Step-by-step explanation:

GIVEN THAT, <ADC=130°

In ADC,

SINCE, DC= AD

SO, <DAC=<ACD= (180-130)/2=50/2=25°

ABC, < ACB =90° (COZ TRIANGLE INSIDE SEMI CIRCLE)

IN QUADRILATERAL,

OPPOSITE ANGLE IS 180°

<ADC+<ABC =180°

130+<ABC=180°

<ABC=50°

ABC ABC,

ACB + ABC + BAC = 180°

90+50+<BAC =180

<BAC=40°

<BAC+<DAC = 40+25 =65°

<ADC=130°

<ACD+<BCA=25+90 =115°

<ABC=50°