For the triangle Below BD=
6
6 CD=
20
20 and BC=
2
109
2
109
solve for sin(B) cos(B) and tan(B) rounded to nearest tenth (note sides not drawn to scale)
To solve for sin(B), cos(B), and tan(B) we can use the following trigonometric formulas:
sin(B) = opposite/hypotenuse
cos(B) = adjacent/hypotenuse
tan(B) = opposite/adjacent
First, let's find the missing side AC using the Pythagorean theorem:
AC^2 = AB^2 + BC^2
AC^2 = 109^2 + 2^2
AC^2 = 11881 + 4
AC^2 = 11885
AC = sqrt(11885)
AC ≈ 109.0
Now we can calculate sin(B), cos(B), and tan(B):
sin(B) = BD/AC = 6/109 ≈ 0.1 (rounded to nearest tenth)
cos(B) = CD/AC = 20/109 ≈ 0.2 (rounded to nearest tenth)
tan(B) = BD/CD = 6/20 = 0.3 (rounded to nearest tenth)
Therefore, sin(B) ≈ 0.1, cos(B) ≈ 0.2, and tan(B) = 0.3.