For the triangle Below BD=

Question

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)

Answer 1

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.