Question

3) In given figure AC perpendicular on BD find value of x. ​

Answer 1

Answer:

In∆abc, angle b+ angle a=90

angle b=50

In∆ deb angle b + angle deb+ angle x=180

50+100+angle x =180

angle x =180-50-100=30 Ans

Answer 2

SOLUTION :-

In ∆ABC,

angle ( A + B + ACB ) = 180° (Angle sum property of a triangle)

40° + B + 90° = 180°

B + 130° = 180°

B = 180° - 130°

B = 50°

In ∆BDE,

angle ( B + BED + D ) = 180° (Angle sum property of ∆)

50° + 100° + x = 180°

150° + x = 180°

x = 180°-150°

x = 30°