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)=

1 answer

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.
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