Question

in a triangle ABC right angled at B. BC = 20. angle A =30. find AC
plz answer correctly if u blabber like something I will report ur ID

Answer 1

Answer:

Sin 30 degree= opp/hypo

1/2=bc/ac

1/2=20/x

x=20/2

x=10

x=ac

ac=10

Answer 2

Answer:

AC = 40/✓3

Step-by-step explanation:

Draw a right angled ∆ABC using given instructions:

Here sec30° = AC/BC

2/✓3 =x/20

40=x✓3

x =40/✓3

By Pythagoras theoram:

(AC)^2 = (AB)^2+ (BC)^2

(40/✓3)^2 = (AB)^2 + (20)^2

1600/3 = (AB)^2 + 400

(AB)^2 = 1600/3 - 400/1

(AB)^2 = 1600 - 1200/3

(AB)^2 = 400/3

AB = ✓400/3

AB = 20/✓3