​Please help
Q. In the given figure, AC$\perp$BD. Find y it $\text{∠BAC=40° and ∠BED=100°}$.

 

In ∆ ABC we have , Angle BAC + Angle ACB + Angle ABC = 180° ( SUM OF ANGLES OF A TRIANGLE) 40°+90°+Angle ABC = 180°, 130°+Angle ABC = 180° , Angle AbC = 50° , In ∆ EBD we have , Angle BED + Angle EBD + y = 180° , 100°+50°+y=180° , 150°+y=180°, y= 180°-150°=30° THEREFORE, y =30° ..

In ΔABC 
∠A + ∠​B + ∠​C = 180⁰ (∠​ sum property)
40⁰ + ∠​B + 90⁰ = 180⁰
130⁰ + ∠​B = 180⁰
∠​B = 180⁰-130⁰
∠​B = 50⁰
In ​ΔEBD
∠E + ∠B + ∠D = 180⁰ (∠​ sum property)​
100⁰ + 50⁰ + y = 180⁰
y = 180⁰-150⁰
y = 30⁰