For the triangle Below BD=1~CD=18 and BC=5\sqrt{13} solve for sin(B) cos(B) and tan(B) rounded to nearest tenth (note sides not drawn to scale)
C
B
D
sin(B)=
cos(B)=
tan(B)=
To solve for sin(B), cos(B), and tan(B), we need to find the length of the remaining side AD first.
Let's use the Pythagorean theorem to find the length of AD:
BD^2 + AD^2 = BA^2
1^2 + AD^2 = (5√13)^2
1 + AD^2 = 25 * 13
AD^2 = 325 - 1
AD^2 = 324
AD = √324
AD = 18
Now we can find sin(B), cos(B), and tan(B):
sin(B) = BD/BA = 1/(1+18) = 1/19 ≈ 0.05
cos(B) = AD/BA = 18/(1+18) = 18/19 ≈ 0.95
tan(B) = BD/AD = 1/18 ≈ 0.06
Therefore, sin(B) ≈ 0.1, cos(B) ≈ 0.9, and tan(B) ≈ 0.1.