Question

Foundation & Olympiad Explorer 5. Triangles and it In the given figure, BCD is a straight line. Find the value of x + y. D y 50° X A B​

Answer 1

Answer:

x - 20 , y-110

Step-by-step explanation:

In ∆ACD :-

ac =ad

angle acd =angle adc

angle cab =50°

angle cad - 40°

so, angle acd = angle adc=(180°-40°) /2

= 70°

In ∆ABD

angle (d+a+b) - 180°

a=90° , d-70,b=x°

= 90°+70°+x′ =180

x° = 20°

angle y + angle acd =180°

y = 110°

Answer 2

Answer:

x+y= 130°

Step-by-step explanation:

cad= 40°(90-50)

cda=dca= 70° (isosceles triangle)

y= 180°-70°= 110° (linear pair)

x= 180-110+50

x= 20°

x+y = 20°+110° = 130°